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I'm wondering as to why I got this physics question wrong, despite it being quite simple in nature. I believe that I missed something while working on the problem.

Basically, the question involves two workers moving a 52 kg crate with a coefficient of friction of 0.52. One pushes at a force of 340 N, while the other pulls with a force of 170 N. The workers push at an angle of 25 degrees.

Credit: Nelson Physics 12

My goal is to find the acceleration of the crate. Since one worker is pushing and the other is pulling on the crate, I thought that it made sense to combine both the pushing and pulling forces together into a single force. Then, I could just subtract the force of friction to find the system's acceleration.

The 340 N push adds to the normal force on the crate (increasing friction) and the 170N subtracts from the normal force on the crate (reducing friction).

applied force - force of friction = mass of system * acceleration
(340cos25 + 170cos25) - (0.52)(9.8*52 + 340sin25 - 170sin25) = 52a
a = 3.07 m/s^2

The textbook's answer is 4.5 m/s^2, which I believe comes by omitting 340sin25:

(340cos25 + 170cos25) - (0.52)(9.8*52 - 170sin25) = 52a
a = 4.5 m/s^2

Why does the book omit the downwards force applied by the first man?

Thanks!

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closed as off-topic by Brandon Enright, ACuriousMind, John Rennie, JamalS, Bernhard Oct 5 '14 at 9:45

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    $\begingroup$ The book would be wrong to omit one of the forces in the system wen computing the response. I suggest you send a note to the publishers (or the author) to point this out. They are normally vey happy when they get an erratum. Chance for pushing another edition, and all the schools have to buy new books... $\endgroup$ – Floris Oct 4 '14 at 16:40
  • $\begingroup$ @Floris Last sentence is the point! $\endgroup$ – an offer can't refuse Oct 5 '14 at 3:24
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The perpendicular force is no longer the weight of the box, you need to consider the force by the two workers in the perpendicular direction as well.

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  • $\begingroup$ I reedited the question to account for that. However, there's still a slight hitch (see above). $\endgroup$ – bubbles Oct 4 '14 at 15:37
  • $\begingroup$ @d.free Maybe you misunderstand pull and push? $\endgroup$ – an offer can't refuse Oct 4 '14 at 15:43
  • $\begingroup$ Push adds to the force of gravity, pull lessens it. That's why I added 340sin25 to the normal force and subtracted 170sin25 from it. However, the book seems to believe that the push doesn't add to the gravity and only the pull does. $\endgroup$ – bubbles Oct 4 '14 at 15:45
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    $\begingroup$ @d.free Maybe the book is not right $\endgroup$ – an offer can't refuse Oct 4 '14 at 15:47
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You have a typo in the above; "9.8*5.2" should read "9.8*52", in two places. But that's apparently just a typo that doesn't reflect your actual calculation; your answer of $3.07 m/s^2$ looks correct. It does look like a mistake in the book.

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  • $\begingroup$ Edited. It doesn't reflect the actual calculation though. $\endgroup$ – bubbles Oct 4 '14 at 16:08
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    $\begingroup$ Well, the answer of 3.1 m/s^2 would look even more correct, since the coefficient of friction is given to only two significant figures... $\endgroup$ – Floris Oct 4 '14 at 16:39

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