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Question

a. Block A as shown in figure, weights 60.0N. The coefficient of static friction between the block and the surface on which it rests is 0.25. The weight w is 12.0 N and the system is in equilibrium. Find the friction force exerted on block A.
b. Find the maximum weight w for which the system will remain in equilibrium.
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Solution

(a) The tension in wire T attached to the hook can be determined as,
Tsin(θ)w
T=wsin(θ)
The force due to static friction on block A is equal to the horizontal component of the
tension.
The expression for frictional force is,
Ff=Tcos(θ)
Substitute T=wsin(θ) in the frictional force expression
Ff=wsin(θ)cos(θ)
=wcot(θ)
Substitute 12 N for w and 45 for θ.
Ff=(12N)cot(45)
=12N
Therefore friction force exerted on block A is 12N.
(b)
From the system to remain in equilibrium force of friction should be eqy=ual to the weight w of the block.
The frictional forceis,
Ff=mAgμ2
Here, mA is mass of A, g is acceleration due to gravity and μs is coefficient
of static friction.
Substitute 60 N for mg and 0.25 for μs.
Ff=(60N)(0.25)
=15N
The force of friction should be equal to the weight w of the block so the weight of the
block is also 15 N.
Therefore, maximum weight w for the system to remain in equilibrium is 15 N.

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