Is the answer key wrong? Physics 12U Dynamics

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The discussion centers on a dynamics problem involving calculating the minimum force required to slide a 265 kg box up a ramp with a 30-degree incline and a coefficient of static friction of 0.45. The original poster's calculations resulted in a force of approximately 2973.7 N, which does not match the answer key's value of 2100 N. Participants noted that the problem's wording suggests a "minimum force," implying that pulling the box rather than pushing it would yield a smaller required force due to reduced normal force and friction. Adjusting the calculations for pulling at an angle of 39 degrees led to a revised estimate of around 2170 N, which is closer to the answer key. The conversation highlights the importance of interpreting the problem correctly and considering the effects of the angle of application on the force required.
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Homework Statement


Hello, I was doing a dynamics question, but my answer didn't match the answer key. I want to see if I am wrong, or the book is. The answer in the book is 2100 N. Question: You are helping a friend move, and you need to load a 265 kg box of books. You slide the box up a ramp, which has an incline of 30 degrees, and a coefficient of static friction of 0.45. You apply he force on the box at an angle of 39 degrees with respect to the ramp. Calculate the minimum force needed to slide the box up the ramp.

Homework Equations


Fnet=ma
Fs=(us)(Fn)
Fg=Mg

The Attempt at a Solution


Fn=(265)(9.8)cos30
=2250 N

Fs=us x Fn
=0.45 x 2250
=1012.5 <---This is the force needed to be overcome just to move the box

Also there is a force of gravity we must overcome
Fgx=(265)(9.8)(sin30)
=1298.5

This force of gravity in the x coordinate also must be overcome, so Fax must be Fgx + Fs
Fax=1012.5 + 1298.5
=2311 N

However this is the force that must be applied in the x coordinate, the force is actually being put on the box at an angle of 39 degrees so:

Fay/sin 39 = 2311/ sin51
Fay=1871.4

Fa=root (1871.4 ^2 + 2311 ^2)
= 2973.7 N

So obviously that doesn't match the answer key, can someone please see if my steps are correct, as well as my question understanding, and if so, is the answer key wrong?
 
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I get same answer as you.

Some general comments:

1) You refer to the "x coordinate" but didn't specify that you have defined that as parallel to the slope.
2) This step...
Alameen Damer said:
However this is the force that must be applied in the x coordinate, the force is actually being put on the box at an angle of 39 degrees so:
Fay/sin 39 = 2311/ sin51
Fay=1871.4
Fa=root (1871.4 ^2 + 2311 ^2)
= 2973.7 N
..seems a bit over complicated. I just wrote..
Fa * Cos(39) = 2310
and solved for Fa

3) It's not clear from the problem statement if the box is pushed or pulled up the ramp. We both assumed that it is pushed. I would expect that pulling it up would give a smaller answer (Normal force and friction is reduced) but I haven't checked if that gives you the book answer.

Edit: I've just read the problem statement again and I note that it says to calculate the "minimium force". That would strongly suggest pulling rather than pushing. Give it a go.
 
Thanks for the reply, so I redid it, using the pull I am getting a force of about 2.17 x 10^3 N. I'm assuming this is close enough to what the book wants.
 
CWatters said:
Edit: I've just read the problem statement again and I note that it says to calculate the "minimium force". That would strongly suggest pulling rather than pushing. Give it a go.
It's not really a question of pull v. push, but of whether the given angle is positive or negative with respect to the ramp angle. But I agree that in practical terms positive will tend to correlate with pulling.
 
Yes that's what I had in mind. Pulling at 39 degrees to the ramp reduces the normal force/friction where as pushing increases it.
 
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