Hello cm2hn, have you tried this video again? In case the Javascript enabling the PDF embedded display was interfering with viewing the video, I have made the PDF into a clickable link instead. I havn't heard reports from other students about difficult viewing videos, so I suspect it's an isolated issue. Please try the video again and let me know if there are still any problems.

I think the answer to this question is wrong. The coordinate of the resultant vector is (-142.2 , 65) which means that it is located in quadrant 2. Since the value of tangent is negative in quadrant 2 it makes sense as well. The angle is 24.56 degrees above the negative x - axis, not positive x-axis.

Hi chaegyunkang, thank you for spotting that, and for the thoughtful analysis. You're quite right! I'll change the written answer, and made a note about the video.
All the best with your studies,
Mr. Dychko

If I am not familiar on the sine law and I am not being instructed in the sine/cosine law, what formulas from the book can be used to arrive at the same solutions?

So... the alternative to using the sine law here (or cosine law in other questions) is to do a lot of work resolving vectors into components, adding those components, then using the Pythagorean Theorem to turn the resultant components into the final resultant vector. The sine and cosine laws are designed to make the solution much, much quicker, and I'm quite sure the time you put into learning them will pay off by avoiding the much larger amount of time dealing with components. Places to start would be https://en.wikipedia.org/wiki/Law_of_sines and https://en.wikipedia.org/wiki/Law_of_cosines. Short term pain for long term gain!
Good luck,
Mr. Dychko

I'm confused on why the last problem did not utilize the Pythagorean theorem to find the actual speed traveled by the swimmer and the actual distance traveled, but for this problem, we did both... I solved the last problem (#46) using the a^2 + b^2 approach and found the angle using the inverse tangent, but my answer was wildly different. Any insight into how to reconcile this in my brain?

Hi brysongrondel, thanks for the question. I understand that something here is confusing, but it isn't clear to me which question/solution you're comparing with, since #46 isn't really comparable to this one... I notice you asked another question, so I'll follow up there.

So, first off, I have to say that I love this site so far. I am kind of struggling with the issues that arise from rounding errors. I am familiar with significant figures, but I am wondering if there is a better rule of thumb. Normally I don't round until I get the final answer, but I got problem #7 wrong. On this problem, I did all of the steps correctly, but I rounded up the times t1 and t2 to 2 sig figs and ended up getting 62 km/h. Is there anything I should know that might be helpful as I prepare for tests and future problem sets?
Thanks!

Hi brysongrondel, thank you very much for the nice feedback, and your question. The general idea is to avoid rounding until the final answer. Rounding values before a final answer is called intermediate rounding error, and causes the final answer to differ as a result of the rounding. So, generally, avoid rounding until the end. It's a topic that calls for a bit of patience since most numbers need rounding eventually, and the answer you're comparing to may have chosen a different number of digits than you, resulting in a different answer than what you obtained despite your following the best practice of keeping "lots" of digits until the end. Different people could have different opinions on what "lots" means. I would say that two additional digits, beyond those that are significant , should be kept with intermediate numbers.
Hope that helps,
Mr. Dychko

Hello, in the part to replace cos45 if I write sq root 2 / 2 (that is what the calculator gives me) instead of 1 / sq root of 2, at the end I guess it will give me a different answer. Right?

Hello cm2hn, thanks for your question. Well, the ultimate proof is always in trying your variation and see the effect, but I can say that $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. If you multiply $\dfrac{1}{\sqrt{2}}$ by "1", you won't change it's value, but if you make "1" look funny as, say, $\dfrac{\sqrt{2}}{\sqrt{2}}$, you'll find that it turns $\dfrac{1}{\sqrt{2}} $ into $\dfrac{\sqrt{2}}{2}$. We know that they're equivalent since you can turn one into the other by multiplying by 1.
All the best,
Mr. Dychko

Hello, thanks for the question. The "y" in the denominator is just a subscript for the "a" (acceleration), to label the acceleration as "the acceleration in the y direction". It is not a separate factor, but maybe I wrote it a bit too big. The equation is $t = \sqrt{\dfrac{2Y}{a_y}}$.

Hello, these are good questions. 365 days / year is a definition, so it's precise to three significant figures. 78 years / lifetime is an estimate, and personal opinion says that being precise to the "ones" place is appropriate, but it's just an opinion. Keep in mind that you never round numbers until the final answer. If you round before, even if a number such as 78 should be 80 since it has one significant figure, suppose, you would still nevertheless do calculations with 78, not 80. Only final answers get rounded, otherwise your calculation has an intermediate rounding error, which is to say that the calculation will be dramatically different only on account of rounding, not because any quantities have actually changed.

Hmmm, that's a thought provoking example. The goal is always to make your calculation as precise as appropriate. Avoiding intermediate rounding error is normally part of that. However, example 1-6 illustrates that if the surrounding values have only one significant figure, as is the case with educated guesses, having only one significant figure in pi is acceptable since, in this context, greater precision isn't called for on account of the other imprecise quantities in the calculation. Think of these significant figure rules as guidelines which are open to interpretation, not as firm rules.

but remember the example (Example 1–6) (How much water is in this lake?) in the book . he is rounded the value of pi to 3
is he rounded for to be 1 SIGFIG or other reason since pi is also in term of (definition) as you mentioned above ?

The actual marry-go-round disk is the total weight - two persons. Which is 560-50 = 510 kg. So, Wouldn't the moment of inertia have to be (2.5)^2(510/2 + 2(25)? I don't understand why you used 560 when calculating the moment of inertia.

Hi theovilous, thanks for the question. What I'm seeing is that the text says

Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 560 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge.

which gives 560kg as the mass of the merry-go-round disk, not including the children. Perhaps you read that differently and interpreted that the children were included in the 560kg?

I am very confused about how the water can evaporate in room temperature. I know that the water cannot reach 100 celsius from our skin but how can water just evaporate in room temp?

Hi chaegyunkang, nice question! It turns out that temperature is an average kinetic energy of the particles of the substance. Water at $100^\circ \textrm{ C}$ has molecules with an average kinetic energy sufficient to turn into gas, so many of them do really quickly, and we observe this as boiling. Water at, say, $20^\circ \textrm{ C}$ also has some molecules with enough kinetic energy to turn into gas, but not very many, so we don't see anything special such as bubbles or steam. Nevertheless those few molecules with enough energy to do so turn into gas. This is what's responsible for evaporation, and eventually the luke warm water will all turn to gas for this reason.

Even ice evaporates! A full ice cube tray in your freezer will eventually have much smaller ice cubes after months since some of the molecules have enough kinetic energy to turn into a gas, even though the average kinetic energy is such that the molecules are in the ice phase.

Hi aquaoasis14, thanks for the question. At the 2:00 minute mark in the video, I multiply by 4 since 4 is the lowest common denominator of the fractions in the equation. Doing this gets rid of the fractions since, as a matter of personal opinion, they're annoying! It would be possible to solve the equation just fine while keeping the fractions, if you prefer. The last term in the equation has 1/2 multiplied by 1/2 (making 1/4), keep in mind, so that's why I needed to multiply by 4 instead of 2.

Hi chaegyunkang, it looks to me like the question is asking for mass per time, which means there needs to be units of mass divided by time, which could be kg / hr, but not just kg.

Hi mbnoroozi, thanks for the question. The final answer I have is positive, but I think you're referring to the negative in the formula for the induced EMF? That negative is there to remind us that the induced EMF causes a current directed such that it produces a magnetic flux in the coil that opposes the change in flux that caused the EMF in the first place. Negative in this case can be thought of as "opposite to", in the same way as a car going in the opposite direction to a car with velocity 30 m/s will have a velocity -30 m/s.

In this problem, can I use the formula, Ffr = μk*Fn? I calculated μk to be .2. So the answer should be .2*60kg*9.8m/s2 = 117.6N. Is this just another way of solving the problem or did I get it wrong? Thank you!!

Hi theovilous, ah, now I understand your approach. Yes, you can do what you're suggesting. I suspect that if you keep lots of digits in your answer for $\mu_k$ (to avoid intermediate rounding error), that your answer will match the answer here exactly to three significant figures. The approach you're suggesting takes more steps, but it's perfectly valid.

Hi theovilous, thanks for the question. While $F_{fr} = \mu_k F_n$ is a correct formula, I'm not sure how you calculated $\mu_k$? It doesn't look possible with the information given, so I think using acceleration is the only method available for solve this one.

Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

Hi samkvertus, thanks for the question. Both the magnitude and directly are actually given here. Direction in this case is "to the left" when the resultant is negative, whereas it's "to the right" when positive. The magnitude is the number when ignoring the negative sign. This is explained more in the video, so consider giving it a second view.

For part (a), I understand how you used the cosine rule to find the unknown vector C. However, I tried to solve the problem similar to what you did in problem 42 and divided the wind velocity into X and Y components. I used to pythagorean theorem to find the resultant vector but was not able to get the right answer. Can you please tell me why it is wrong to use this method.

Hi chaegyunkang, thanks for the question. The method you describe works fine, so your sleuthing should look at how you implemented it. Pay careful attention to whether you subtracted the y-component of the velocity of the air with respect to the ground from the velocity of the plane with respect to the air (instead of adding it). Knowing that $\cos(45) = \dfrac{1}{\sqrt{2}}$, here's what the work will look like in your calculator:
$\Big(688 - \dfrac{90}{\sqrt{2}} \Big)^2 + \Big(\dfrac{90}{\sqrt{2}}\Big)^2 = \textrm{ some number}$
$\sqrt{ \textrm{ some number}} = 627.5953...$ which is the correct magnitude, and I'll leave the angle for you.

Hello, I got a different answer for this question. I divided all the vectors into x and y components and used pythagorean theorem at the end to find the resultant vector. My answer is 22.53 km ( 22.53 degrees North of East ).

Hi chaegyunkang, that's a fair question that other's have had as well. I'll paste my response to another student's similar point: "this problem is meant to check a students ability to plot vectors on graph paper in the correct direction, and add them together. If you reduced the vectors to components, then added the components together, I would bet that your answer is more precise than mine. If you have an answer that is a some length more than 15 m and less than 30 m, then consider your answer correct. I followed the text book instructions to add them graphically, rather than use a calculator. There is a human error in plotting the length and direction for each arrow, and this happens three times, so the errors are compounding. Actually, the error compounds four times when you consider the error in the resultant, so don't be surprised of your answer differs from what's shown. The thing to learn from this video is how to plot vectors, and what it means, graphically speaking, to add three of them together."

Hi missedinger, that's a totally fair question, which I've addressed in the FAQ: It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$.

The only thing that matters is that you're consistent with your coordinate system. This is to say that everything pointing in a particular direction needs the same sign (be it positive or negative doesn't matter, just that they're all the same), and things pointing in the opposite direction have the opposite sign. So, yes, you could make the capsule velocity negative, keeping in mind that this now means the force exerted on the capsule by the astronaut would also be negative in that coordinate system, since it would be directed to the right just the same as the now negative capsule velocity.

Hi cm2hn, we could have used the space capsule final speed in part b), as you suggest, but that's taking a bit of a risk since we calculated the space capsule speed ourselves in part a), and we might have made an error in that calculation. Using the astronaut speed provided by the question is a safer approach, but outside of that, either speed is fine.

When a question doesn't specify an initial speed we can assume it's 0? I'm used to seeing them specify "starting from rest". This is only true with an initial speed of 0.

Hi idan, thank you for the question. When a question doesn't specify an initial speed, and doesn't imply it by saying something like "lifted off the ground" or saying something else which could be interpreted as an initial speed, then you really don't know what the initial speed is. You can't assume it is zero. In this question, the initial speed doesn't actually matter until you get to part e) where it's asking for the final speed. In part e) it finally says "it started from rest". For parts a) through d), the force needed to impart an acceleration of $0.160 \textrm{ g}$ is the same regardless of what speed it's going initially. Only the acceleration is relevant. For example, consider that the force needed to keep the load at rest would be the same as the force needed to make the load rise at a constant speed. The constant speed doesn't matter so far as the force is concerned.

## Giancoli 7th Edition, Chapter 4, Problem 22

By chaegyunkang on Sun, 07/01/2018 - 15:55where is the normal force for part a) ?

Oh, I realized that free body diagram only shows the forces acting on the object, not on other objects....

Yep, you got it! :)

## Giancoli 7th Edition, Chapter 19, Problem 20

By cm2hn on Sat, 06/30/2018 - 21:13Hello, I was trying yo watch the video, but it is unable to play. Thanks.

Hello cm2hn, have you tried this video again? In case the Javascript enabling the PDF embedded display was interfering with viewing the video, I have made the PDF into a clickable link instead. I havn't heard reports from other students about difficult viewing videos, so I suspect it's an isolated issue. Please try the video again and let me know if there are still any problems.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 16

By chaegyunkang on Wed, 06/27/2018 - 20:21I think the answer to this question is wrong. The coordinate of the resultant vector is (-142.2 , 65) which means that it is located in quadrant 2. Since the value of tangent is negative in quadrant 2 it makes sense as well. The angle is 24.56 degrees above the negative x - axis, not positive x-axis.

Hi chaegyunkang, thank you for spotting that, and for the thoughtful analysis. You're quite right! I'll change the written answer, and made a note about the video.

All the best with your studies,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 48

By brysongrondel on Sat, 06/02/2018 - 20:49If I am not familiar on the sine law and I am not being instructed in the sine/cosine law, what formulas from the book can be used to arrive at the same solutions?

So... the alternative to using the sine law here (or cosine law in other questions) is to do a lot of work resolving vectors into components, adding those components, then using the Pythagorean Theorem to turn the resultant components into the final resultant vector. The sine and cosine laws are designed to make the solution much, much quicker, and I'm quite sure the time you put into learning them will pay off by avoiding the much larger amount of time dealing with components. Places to start would be https://en.wikipedia.org/wiki/Law_of_sines and https://en.wikipedia.org/wiki/Law_of_cosines. Short term pain for long term gain!

Good luck,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 47

By brysongrondel on Sat, 06/02/2018 - 19:58I'm confused on why the last problem did not utilize the Pythagorean theorem to find the actual speed traveled by the swimmer and the actual distance traveled, but for this problem, we did both... I solved the last problem (#46) using the a^2 + b^2 approach and found the angle using the inverse tangent, but my answer was wildly different. Any insight into how to reconcile this in my brain?

Hi brysongrondel, thanks for the question. I understand that something here is confusing, but it isn't clear to me which question/solution you're comparing with, since #46 isn't really comparable to this one... I notice you asked another question, so I'll follow up there.

## Giancoli 7th Edition, Chapter 2, Problem 12

By brysongrondel on Mon, 05/28/2018 - 16:30So, first off, I have to say that I love this site so far. I am kind of struggling with the issues that arise from rounding errors. I am familiar with significant figures, but I am wondering if there is a better rule of thumb. Normally I don't round until I get the final answer, but I got problem #7 wrong. On this problem, I did all of the steps correctly, but I rounded up the times t1 and t2 to 2 sig figs and ended up getting 62 km/h. Is there anything I should know that might be helpful as I prepare for tests and future problem sets?

Thanks!

Hi brysongrondel, thank you very much for the nice feedback, and your question. The general idea is to avoid rounding until the final answer. Rounding values before a final answer is called

intermediate rounding error, and causes the final answer to differ as a result of the rounding. So, generally, avoid rounding until the end. It's a topic that calls for a bit of patience since most numbers need rounding eventually, and the answer you're comparing to may have chosen a different number of digits than you, resulting in a different answer than what you obtained despite your following the best practice of keeping "lots" of digits until the end. Different people could have different opinions on what "lots" means. I would say thattwo additional digits, beyond those that are significant, should be kept with intermediate numbers.Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 16, Problem 29

By cm2hn on Mon, 05/28/2018 - 12:12Hello, in the part to replace cos45 if I write sq root 2 / 2 (that is what the calculator gives me) instead of 1 / sq root of 2, at the end I guess it will give me a different answer. Right?

Hello cm2hn, thanks for your question. Well, the ultimate proof is always in trying your variation and see the effect, but I can say that $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. If you multiply $\dfrac{1}{\sqrt{2}}$ by "1", you won't change it's value, but if you make "1" look funny as, say, $\dfrac{\sqrt{2}}{\sqrt{2}}$, you'll find that it turns $\dfrac{1}{\sqrt{2}} $ into $\dfrac{\sqrt{2}}{2}$. We know that they're equivalent since you can turn one into the other by multiplying by 1.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 7, Problem 34

By rodri100172 on Thu, 05/17/2018 - 18:51Thanks bro! you're a lifesaver! Say hi to Brian for me.

## Giancoli 7th Edition, Chapter 3, Problem 17

By nguyen112212 on Tue, 05/15/2018 - 19:51Hi Sr. I am wondering what happened to the other "Y" when you were solving for time. It's the part where t = square root of 2Y/AY.

Hello, thanks for the question. The "y" in the denominator is just a subscript for the "a" (acceleration), to label the acceleration as "the acceleration in the y direction". It is not a separate factor, but maybe I wrote it a bit too big. The equation is $t = \sqrt{\dfrac{2Y}{a_y}}$.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 5, Problem 43

By pathak_s on Sun, 05/06/2018 - 17:44Hey why doesn't the equation for velocity (2*pi*r)/T work for this question as well?

## Giancoli 7th Edition, Chapter 19, Problem 24

By aquaoasis14 on Wed, 04/18/2018 - 18:01why did the R disappear finding the currents to the after problem?(the last problem)

## Giancoli 7th Edition, Chapter 22, Problem 14

By william.g.parker on Wed, 04/04/2018 - 15:46You wrote 0.314 as the answer for B instead of 0.341

Fixed! Thank you!

## Giancoli 7th Edition, Chapter 1, Problem 28

By abgoooor201596 on Tue, 04/03/2018 - 02:10Why your not estimating the Numbers 362 and 76 to one SIGFIG. ?

Hello, these are good questions. 365 days / year is a definition, so it's precise to three significant figures. 78 years / lifetime is an estimate, and personal opinion says that being precise to the "ones" place is appropriate, but it's just an opinion. Keep in mind that you never round numbers until the

final answer. If you round before, even if a number such as 78 should be 80 since it has one significant figure, suppose, you would still nevertheless do calculations with 78, not 80. Only final answers get rounded, otherwise your calculation has anintermediate rounding error, which is to say that the calculation will be dramatically different only on account of rounding, not because any quantities have actually changed.Hope that helps,

Mr. Dychko

Hmmm, that's a thought provoking example. The goal is always to make your calculation as precise as appropriate. Avoiding intermediate rounding error is normally part of that. However, example 1-6 illustrates that if the surrounding values have only one significant figure, as is the case with educated guesses, having only one significant figure in pi is acceptable since, in this context, greater precision isn't called for on account of the other imprecise quantities in the calculation. Think of these significant figure rules as guidelines which are open to interpretation, not as firm rules.

All the best,

Mr. Dychko

Ok That is good idea

but remember the example (Example 1–6) (How much water is in this lake?) in the book . he is rounded the value of pi to 3

is he rounded for to be 1 SIGFIG or other reason since pi is also in term of (definition) as you mentioned above ?

Thanks

## Giancoli 7th Edition, Chapter 1, Problem 25

By abgoooor201596 on Tue, 04/03/2018 - 01:28hi I've a question

How we gona apply this idea with this Number 3.0*10^0

Thank You

Hi abgoooor201596, $3.0 \times 10^0$ has 2 significant figures.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 8, Problem 41

By theovilous on Sat, 03/31/2018 - 10:31The actual marry-go-round disk is the total weight - two persons. Which is 560-50 = 510 kg. So, Wouldn't the moment of inertia have to be (2.5)^2(510/2 + 2(25)? I don't understand why you used 560 when calculating the moment of inertia.

Hi theovilous, thanks for the question. What I'm seeing is that the text says

which gives 560kg as the mass of the merry-go-round disk, not including the children. Perhaps you read that differently and interpreted that the children were included in the 560kg?

Best wishes,

Mr. Dychko

I read the question wrong! thanks for pointing that out! Thank you so much as always!!

## Giancoli 7th Edition, Chapter 14, Problem 26

By chaegyunkang on Wed, 03/07/2018 - 18:41I am very confused about how the water can evaporate in room temperature. I know that the water cannot reach 100 celsius from our skin but how can water just evaporate in room temp?

Hi chaegyunkang, nice question! It turns out that temperature is an

averagekinetic energy of the particles of the substance. Water at $100^\circ \textrm{ C}$ has molecules with an average kinetic energy sufficient to turn into gas, so many of them do really quickly, and we observe this as boiling. Water at, say, $20^\circ \textrm{ C}$alsohas some molecules with enough kinetic energy to turn into gas, but not very many, so we don't see anything special such as bubbles or steam. Nevertheless those few molecules with enough energy to do so turn into gas. This is what's responsible for evaporation, and eventually the luke warm water will all turn to gas for this reason.Even ice evaporates! A full ice cube tray in your freezer will eventually have much smaller ice cubes after months since some of the molecules have enough kinetic energy to turn into a gas, even though the average kinetic energy is such that the molecules are in the ice phase.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 8, Problem 58

By aquaoasis14 on Sun, 03/04/2018 - 19:42why did you multiply it by 4?

Hi aquaoasis14, thanks for the question. At the 2:00 minute mark in the video, I multiply by 4 since 4 is the lowest common denominator of the fractions in the equation. Doing this gets rid of the fractions since, as a matter of personal opinion, they're annoying! It would be possible to solve the equation just fine while keeping the fractions, if you prefer. The last term in the equation has 1/2 multiplied by 1/2 (making 1/4), keep in mind, so that's why I needed to multiply by 4 instead of 2.

Hope that helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 14, Problem 7

By chaegyunkang on Thu, 03/01/2018 - 16:50If the question asked for amount of water, is it okay to end with kg of water? Thank You.

Hi chaegyunkang, it looks to me like the question is asking for mass per time, which means there needs to be units of mass divided by time, which could be kg / hr, but not just kg.

All the best,

Mr. Dychko

## Giancoli 6th Edition, Chapter 21, Problem 16

By mbnoroozi on Sun, 02/25/2018 - 14:51Hi Dr. Dychko - I might be missing something obvious but can you explain why the final answer is negative?

Thanks!

Hi mbnoroozi, thanks for the question. The final answer I have is positive, but I think you're referring to the negative in the formula for the induced EMF? That negative is there to remind us that the induced EMF causes a current directed such that it produces a magnetic flux in the coil that opposes the change in flux that caused the EMF in the first place. Negative in this case can be thought of as "opposite to", in the same way as a car going in the opposite direction to a car with velocity 30 m/s will have a velocity -30 m/s.

Hope this helps,

Mr. Dychko

## Giancoli 7th Edition, Chapter 4, Problem 51

By theovilous on Fri, 02/16/2018 - 07:47In this problem, can I use the formula, Ffr = μk*Fn? I calculated μk to be .2. So the answer should be .2*60kg*9.8m/s2 = 117.6N. Is this just another way of solving the problem or did I get it wrong? Thank you!!

Hi theovilous, ah, now I understand your approach. Yes, you can do what you're suggesting. I suspect that if you keep lots of digits in your answer for $\mu_k$ (to avoid intermediate rounding error), that your answer will match the answer here exactly to three significant figures. The approach you're suggesting takes more steps, but it's perfectly valid.

I'm glad you're enjoying the solutions!

All the best,

Mr. Dychko

Hi theovilous, thanks for the question. While $F_{fr} = \mu_k F_n$ is a correct formula, I'm not sure how you calculated $\mu_k$? It doesn't look possible with the information given, so I think using acceleration is the only method available for solve this one.

All the best,

Mr. Dychko

I considered retarding force as friction. So I set the formula Ffr = ma --> μkmg=ma --> μk = a/g --> μk = 2/9.8 = .2.

So based on the formula, Ffr=μkFn, Ffr=μkmg = .2 * 60kg * 9.8m/s2 = 117.6N.

Does this approach work as well? my result is very close to your answer which is 120N.

Thank you for your quick response! I am learning a lot from you!

## Giancoli 7th Edition, Chapter 8, Problem 59

By choianchoi on Sun, 01/21/2018 - 17:02Why is kinetic energy not accounted just before it hits the ground? Also, is the velocity of the center of mass the same with the end of the pole?

Hi choianchoi, thanks for the question. Let's keep in mind that one end of the pole stays in contact with the ground. It's a pole falling over, in other words. The end touching the ground has velocity zero in that case, at all times. The other end of the pole, on the other hand, has the maximum velocity of any point on the pole. The end initially in the air has a higher linear velocity than the center of mass. Kinetic energy is accounted for, and it's the strategy for calculating the velocity of the end of the pole, but the kinetic energy formula doesn't look like what you're used to from linear problems. We don't use $\dfrac{1}{2}mv^2$. Instead, we use rotational kinetic energy $\dfrac{1}{2}I\omega^2$ instead, and use that to figure out the rotational velocity, which is then used to find the linear velocity of the end of the pole.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 24

By samkvertus on Thu, 01/18/2018 - 20:54amazing

Thanks!

## Giancoli 7th Edition, Chapter 3, Problem 7

By samkvertus on Fri, 01/12/2018 - 11:20what about the last part of the problem that asked to find the magnitude and direction ?

Hi samkvertus, thanks for the question. Both the magnitude and directly are actually given here. Direction in this case is "to the left" when the resultant is negative, whereas it's "to the right" when positive. The magnitude is the number when ignoring the negative sign. This is explained more in the video, so consider giving it a second view.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 3, Problem 44

By chaegyunkang on Wed, 01/10/2018 - 08:58Dear Mr. Dychoko,

For part (a), I understand how you used the cosine rule to find the unknown vector C. However, I tried to solve the problem similar to what you did in problem 42 and divided the wind velocity into X and Y components. I used to pythagorean theorem to find the resultant vector but was not able to get the right answer. Can you please tell me why it is wrong to use this method.

Thank You

Hi chaegyunkang, thanks for the question. The method you describe works fine, so your sleuthing should look at how you implemented it. Pay careful attention to whether you subtracted the y-component of the velocity of the air with respect to the ground from the velocity of the plane with respect to the air (instead of adding it). Knowing that $\cos(45) = \dfrac{1}{\sqrt{2}}$, here's what the work will look like in your calculator:

$\Big(688 - \dfrac{90}{\sqrt{2}} \Big)^2 + \Big(\dfrac{90}{\sqrt{2}}\Big)^2 = \textrm{ some number}$

$\sqrt{ \textrm{ some number}} = 627.5953...$ which is the correct magnitude, and I'll leave the angle for you.

Hope this helps,

Mr. Dychko

My mistake was adding the vectors. Thank you so much.

## Giancoli 7th Edition, Chapter 3, Problem 4

By chaegyunkang on Sun, 12/31/2017 - 10:07Hello, I got a different answer for this question. I divided all the vectors into x and y components and used pythagorean theorem at the end to find the resultant vector. My answer is 22.53 km ( 22.53 degrees North of East ).

Hi chaegyunkang, that's a fair question that other's have had as well. I'll paste my response to another student's similar point: "this problem is meant to check a students ability to plot vectors on graph paper in the correct direction, and add them together. If you reduced the vectors to components, then added the components together, I would bet that your answer is more precise than mine. If you have an answer that is a some length more than 15 m and less than 30 m, then consider your answer correct. I followed the text book instructions to add them graphically, rather than use a calculator. There is a human error in plotting the length and direction for each arrow, and this happens three times, so the errors are compounding. Actually, the error compounds four times when you consider the error in the resultant, so don't be surprised of your answer differs from what's shown. The thing to learn from this video is how to plot vectors, and what it means, graphically speaking, to add three of them together."

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 6, Problem 3

By missedinger on Sat, 12/02/2017 - 21:05wondering why you stop @ question 71 for solutions, when the text books has 94 questions?

Hi missedinger, that's a totally fair question, which I've addressed in the FAQ: It would be enormously time consuming to also answer all the "General Problems" so I'm limiting coverage only to the regular "Problems", of which there are still more than 1700. If you need help with a "General Problem", my suggestion would be to try and find a regular "Problem" that is similar to the "General Problem", and see if you can apply the same problem solving technique.

All the best,

Mr. Dychko

## Giancoli 7th Edition, Chapter 10, Problem 9

By jr59j on Fri, 12/01/2017 - 11:55Why is it 1/2 of m*g

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$.

Cheers,

Mr. Dychko

## Giancoli 7th Edition, Chapter 7, Problem 19

By cm2hn on Mon, 11/20/2017 - 20:20Why the speed of astronaut is negative? Would I get the same result if I use the capsule as the negative value?

The only thing that matters is that you're consistent with your coordinate system. This is to say that everything pointing in a particular direction needs the same sign (be it positive or negative doesn't matter, just that they're all the same), and things pointing in the opposite direction have the opposite sign. So, yes, you could make the capsule velocity negative, keeping in mind that this now means the force exerted on the capsule by the astronaut would also be negative in that coordinate system, since it would be directed to the right just the same as the now negative capsule velocity.

## Giancoli 7th Edition, Chapter 7, Problem 19

By cm2hn on Mon, 11/20/2017 - 20:06Why average force is based on ma*vfa and not ms*vfs?

Hi cm2hn, we could have used the space capsule final speed in part b), as you suggest, but that's taking a bit of a risk since we calculated the space capsule speed ourselves in part a), and we might have made an error in that calculation. Using the astronaut speed provided by the question is a safer approach, but outside of that, either speed is fine.

## Giancoli 7th Edition, Chapter 6, Problem 25

By idan on Wed, 11/15/2017 - 17:49When a question doesn't specify an initial speed we can assume it's 0? I'm used to seeing them specify "starting from rest". This is only true with an initial speed of 0.

Hi idan, thank you for the question. When a question doesn't specify an initial speed, and doesn't imply it by saying something like "lifted off the ground" or saying something else which could be interpreted as an initial speed, then you really don't know what the initial speed is. You can't assume it is zero. In this question, the initial speed doesn't actually matter until you get to part e) where it's asking for the final speed. In part e) it finally says "it started from rest". For parts a) through d), the force needed to impart an acceleration of $0.160 \textrm{ g}$ is the same regardless of what speed it's going initially. Only the acceleration is relevant. For example, consider that the force needed to keep the load at rest would be the same as the force needed to make the load rise at a constant speed. The constant speed doesn't matter so far as the force is concerned.

Hope this helps,

Mr. Dychko