Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
4
Dynamics: Newton's Laws of Motion
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4-4 to 4-6: Newton's Laws, Gravitational Force, Normal Force
4-7: Newton's Laws and Vectors
4-8: Newton's Laws with Friction; Inclines

Problem 27
A
FB=6.9×103FA+FB=8.9×103NF_B = 6.9 \times 10^3 \textrm{N } F_A+F_B = 8.9 \times 10^3 \textrm{N}
Giancoli 6th Edition, Chapter 4, Problem 27 solution video poster
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VIDEO TRANSCRIPT

These snow cats are towing this housing unit and our job is to figure out what it is the force exerted by the snow cat b, force b. We’re told that the resultant of these two forces is going to be along this vertical axis. So from that we know that the X-component of force b which I have drawn here is going to equal the X-component of force a. In order for the resultant to be along this vertical axis these two X-components have to balance. So let’s say that algebraically we have: ‘Fbx’ equals ‘Fax’ and then we know that ‘Fbx’ equals ‘Fa’ sine fifty, using sine because in this ‘Fa’ force triangle the ‘Fax’ is the opposite leg, opposite to the fifty degrees and so we use sine to get that. We also know that ‘Fbx’ equals ‘Fb’ times sine thirty, now I am looking at the green ‘Fb’ triangle and ‘Fbx’ is the opposite leg of this triangle so ‘Fbx’ is ‘Fb’ sine thirty. We have two expressions of ‘Fbx’ and so they must themselves be equal. So we know that ‘Fb’ is ‘Fbx’ divided by sine thirty. Divide both sides by sine thirty here and then we’ll make a substitution from equation one for ‘Fbx’. So we’ll write: ‘Fb’ is ‘Fa’ sine fifty divided by sine thirty. So we started by saying the X-components are equal then we said ‘Fax’ is the opposite leg of the ‘Fa’ triangle so we use sine fifty times ‘Fa’ to get it. Looking at force triangle for force b and also the opposite leg also leg we use sine thirty in the ‘Fb’ triangle. And solving for ‘Fb’ from this equation two and then substituting from equation one for ‘Fbx’: ‘Fb’ equals four thousand five hundred newtons times sine fifty divided by sine thirty which is six point nine times ten raised to power three newtons. Since the X-components cancelled in this picture, our job now is to find the resultant force and it will be the Y-components of both forces added together since the X-components just cancel. The resultant force of ‘Fa’ plus ‘Fb’ is going to be: ‘Fay’ plus ‘Fby’ which is ‘Fa’ cosine fifty plus ‘Fb’ cosine thirty using the adjacent leg of each of this force triangles, cosine in both cases., four thousand five hundred times cosine fifty plus six thousand eight hundred and ninety four newtons which is the unrounded answer to the first part times cosine thirty and this gives eight point nine times ten raised to power three newtons vertically upwards is the resultant force.

COMMENTS
By emmarubin on Sun, 12/30/2012 - 5:50 PM

if fbx was NOT equal to fax there would be a net horizontal force and the snowcat would not be moving straight in the y direction (it would be moving in the positive x direction if fbx greater than fax and would be moving in the negative x direction if fax greater than fbx)

By monkboy11 on Tue, 5/24/2011 - 5:19 AM

Thx for the explanation. I'm having a bit of trouble wrapping my head around why FAx and FBx are equal to each other. Is is b/c there is no force horizontally and therefore the x-component is constant? Would you mind clarifying? Thanks!!

By jcbristol on Mon, 9/23/2013 - 8:22 PM

I'm having trouble with the substitution 4:32. The value of FAX is 6,894N; meaning, the force in the positive x direction is 6,894N. Why is this value substituted at 4:32? Wouldn't this substitution need the Hypotenuse of the resultant forces from FAX and FAY. Cosine(x)=adjacent/hypotenuse?

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