Two masses A and B of 15 kg and 10 kg are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and block is 0.4 . The minimum mass of C , that may be placed on A to prevent it from moving is :

Question

Two masses A and B of 15 kg and 10 kg are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and block is 0.4 . The minimum mass of C , that may be placed on A to prevent it from moving is : (A)  10 kg  (B)  5 kg  (C) zero (D)  15 kg

Tardigrade Admin · Accepted Answer

For blocks to be stationary, net force on each is zero.
The following free body diagram shows the various forces acting on the system. Let

m

be the minimum mass of block

C

and

f_{s}

be the maximum value of static friction

For block A :

R = (m + m_{A}) g, f_{s} = T

∴ μ (m + m_{A}) g = T \dots (i)

For block B :

T = m_{B} g \dots (ii)

From Eqs. (i) and (ii), we get

m = \frac{m _{B} - μ m _{A}}{μ}

m = \frac{10 - 0.4 \times 15}{0.4} = 10 k g

$10 k g$

$5 k g$

zero

$15 k g$

Q. Two masses A and B of 15kg and 10kg are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and block is 0.4 . The minimum mass of C , that may be placed on A to prevent it from moving is :

10kg

5kg

zero

15kg

$10 k g$

$5 k g$

$15 k g$