Consider the figure below. (Let w1 = 135 N and w2 = 46.5 N.) (a) What is the min

Question

Consider the figure below. (Let w1 = 135 N and w2 = 46.5 N.)

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

(b) What coefficient of static friction between the 135-N block and the table ensures equilibrium?

(c) If the coefficient of kinetic friction between the 135-N block and the table is 0.172, what hanging weight should replace the 46.5-N weight to allow the system to move at a constant speed once it is set in motion?

PLEASE don't solve for other numbers...use the one's provided and if you could explain it along the way that'd be awesome :) thanks in advance.

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 46.5N due to the tension produce in the string due to w2. so, F(net) =f* (tangential force) - f ( frictional force) 0 =46.5 - f f = 46.5 N ( b ). The static friction coefficient (µ) 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) µ = F /N µ = 46.5 / 135 µ = 0.3444 ( 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 23.22 N because, kinetic friction =µ (f)*135 =0.172*135 =23.22 N now if we want the body to slide with constant speed the tension produce in the rope should be of 23.22 N i.e. the applied force on the right side should be 23.22 N, then and then the net force on the system should be 0 N.

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