To move a large crate across a rough floor, you push on it with a force F at an angle of 21°, below the horizontal, as shown in the figure. Find the acceleration of the crate, given that the mass of the crate is m = 27 kg, the applied force is 327 N and the coefficient of kinetic friction between the crate and the floor is 0.66.

Question

anonymous · Answer

I have already tried two ways:
1. using the equation F(net) = Fcosθ - µk x [mg + Fsinθ] then dividing by 27 kg. With this I got an acceleration of 2.68 m/s^2
2. I added all the forces to get F(net)
Ffriction= µkN = -161.57N
Fapplied= 327N
Fgravity= µkmgsinθ= 57.9N
with a total Fnet= 223.33N
Then divided by 27kg to get 8.27 m/s^2

I have on more chance to get the correct answer for this problem and I am so stuck. Any help is appreciated!

anonymous · Answer

You are combining vertical and horizontal forces in both approaches. You need to split them up. You should have a Fnet for each.

anonymous · Answer

Do you have a free body diagram?

anonymous · Answer

Ffriction= µkNcosθ= -163.20N
F= applied*cosθ= 305.28N
Fgravity= mgsinθ= 94.92N
Am i on the right track?

anonymous · Answer

The friction and applied forces are correct. Fgravity isn't. The box is on a flat surface, so Fgravity = mg

anonymous · Answer

Free body diagram
Start by summing all the forces in the horizontal direction. Since the box is only moving horizontally, the acceleration only has a horizontal component.
Fnet_x = ma
Since the box isn't moving vertically, Fnet_y = 0.
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