Question
Asked Oct 10, 2019

 A horizontal force, F1 = 75 N, and a force, F2 = 16.7 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 4.6 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.

F1 = 75 N
F2 = 16.7 N
m = 4.6 kg

a.) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°.

b.) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if θ = 30°.

F.
m
Otheexpertta.com
k
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F. m Otheexpertta.com k

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Step 1

a)

The equation for the normal force N would be

N mg F sin30°
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N mg F sin30°

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Step 2

Substitute 4.6 kg for m, 9.8m/s2 for g, 16.7 N for F2 and solve for N as,

N =(4.6kg)(9.8m/s) + (16.7 sin 30°) N
- 45.08 N+8.35N
- 53.43 N
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N =(4.6kg)(9.8m/s) + (16.7 sin 30°) N - 45.08 N+8.35N - 53.43 N

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Step 3

b)

The magnitude of frictional force ...

f = N
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f = N

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