An 8.00 kg block of steel is at rest on a horizontal table. The coefficient of s

Question

An 8.00 kg block of steel is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. A force is to be applied to the block. What is the magnitude of that applied force if it puts the block on the verge of sliding when the force is directed in the following directions? (a) horizontally
N

(b) upward at 60.0° from the horizontal
N

(c) downward at 60.0° from the horizontal
N (a) horizontally
N

(b) upward at 60.0° from the horizontal
N

(c) downward at 60.0° from the horizontal
N

Explanation / Answer

The mass of the block m = 8.0kg the coefficient of friction µ = 0.5 (a) When the force applied horizontaly Fnet = F - f = 0 therefore the applied force F = µmg = (0.5)(8)(9.8) = 39.2N (b) Upwards at 60
Fnet = Fcos? - µ(mg - Fsin?) = 0 Fcos? = µmg - µFsin? F (cos? + µsin?) = µmg therefore F = µmg / [ cos? + µsin?] = (0.5)(8)(9.8) / [ cos60 + (0.5)sin60)] = 42N
(c) when the force applied downwards
F = µmg / [ cos? - µsin?] = (0.5)(8)(9.8) / [ cos60 - (0.5)sin60)] = 585N
F = µmg / [ cos? - µsin?] = (0.5)(8)(9.8) / [ cos60 - (0.5)sin60)] = 585N = (0.5)(8)(9.8) / [ cos60 - (0.5)sin60)]

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