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Chapter 7: Q. 7 (page 177)

Block B in the figure rests on a surface for which the static and kinetic coefficients of friction are 0.60and 0.40respectively. The ropes are massless.

What is the maximum mass of block A for which the system remains in equilibrium?

Short Answer

Expert verified

The maximum mass of the block A for which the system remains in equilibrium is12kg.

Step by step solution

01

Part (a) Step 1 : Given information

Static friction coefficient 渭s=0.60,kinetic friction coefficient 渭k=0.40,mass of blockmB=20kg.

02

Part (a) Step 2 : Calculation

Here, the tension in the rope is the same at all points. Block B must not slide to be in equilibrium.

That means that the tension T in the rope must not exceed the maximum static friction force acting on B.

T=渭s脳mB脳gT=0.6脳20kg脳9.8m/s2T=117.6N

Tension in the rope at 450 is T . Also, equilibrium requires that block A not to move, so

localid="1650266822568" Tsin45=mAgT=2mAgm蚂=T/2gTcos45=117.6NT=2脳117.6Nm蚂=2脳117.6N2脳g117.6N/9.8m/s2m蚂=12kg

03

Part (a) Step 3 : Final answer

The maximum mass of the block for which the system remains in equilibrium is12kg.

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