Question

Two masses a and b of 10 kg and 5 kg respectively are connected with a string passing over frictionless pulley fixed at the corner of a table as shown. the coefficient of static friction of a with tabel is 0.2. the minimum mass of c that must be placed on a to prevent it from moving i

Answer 1

hope this answer will help you

Answer 2

Hey buddy,

◆ Answer-
mc = 15 kg

◆ Explaination-
# Given-
ma = 10 kg
mb = 5 kg
mc = ?
μs = 0.2

# Solution-
Tension in the string is
T = mb.g

To prevent mass a from moving, mass c must be placed so that total force of friction counteracts tension in string.
F = μs.N = T
μs(ma+mc)g = mb.g

Substituting values-
0.2(10+mc)g = 5g
10 + mc = 25
mc = 15 kg

Minimum mass c must be placed should be 15 kg.

Hope it helps you...