Consider the figure below. (Let w 1 = 120 N and w 2 = 44.5 N.) (a) What is the m

Question

Consider the figure below. (Let w1 = 120 N and w2 = 44.5 N.)

(a) What is the minimum force of friction required to hold the system of the figure above in equilibrium?


(b) What coefficient of static friction between the 120-N block and the table ensures equilibrium? (Enter the minimum acceptable coefficient of friction.)


(c) If the coefficient of kinetic friction between the 120-N block and the table is 0.185, what hanging weight should replace the 44.5-N weight to allow the system to move at a constant speed once it is set in motion?

Explanation / Answer

The minimum force required to hold the system in equillibrium, frictional force acting on the body should be equal to the tangential force acting on it. The tangential force acting on the body 1 is 44.5N due to the tension produce in the string due to w2.
so, F(net) =f* (tangential force) - f ( frictional force)
0 =44.5 - f
f = 44.5 N
( b ).
The static friction coefficient (?) between two solid surfaces is defined as the ratio of the tangential force (F) required to produce sliding divided by the normal force between the surfaces (N)

? = F /N
? = 44.5 / 120
? = 0.3783

( c ).
To allow the system to move at a constant speed once it is set in motion, the the block 2 should be replaced by a block of weight 22.2 N
because,
kinetic friction =? (f)*10
=0.185*120
=22.2 N
now if we want the body to slide with constant speed the tension produce in the rope should be of 22.2 N i.e. the applied force on the right side should be 22.2 N, then and then the net force on the system should be 0 N.

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