The 2.0 kg wood box in the figure(Figure 1) slides down a vertical wood wall while you push on it at a 45 ∘ angle. The coefficient of kinetic friction of wood on wood is μk = 0.200.

What magnitude of force should you apply to cause the box to slide down at a constant speed?


Fnet(y) = Fp(y) - (2.0)(9.8) = 0
Fp = N (μk)
Fp = N (0.200)
Fp = (9.8)(2.0)(0.200)
Fp = 3.92

Fp(y) = 3.92 (Sin(45))
Fp(y) = 2.77

Fnet(y) = 2.77 - (2.0)(9.8)
Fnet(y) = 16 N

The answer is supposed to be 23. What am I doing wrong?

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  1. If I read the problem, you are also pushing upwards Fcos45, as well as inwards.
    in the vertical:
    F= 19.6/(0.5656)=34.7
    Now if you are pushing downward at 45deg, then
    2*9.8 F(-.707-.2*.707)=0
    F=19.6/( 0.848)=23N
    so it I guess has the drawing with the force downward

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