Question

Two masses of 20 kg and 10 kg are connected by
a light string passing over a frictionless pulley
The coefficient of friction between 20 kg mass
and the table is 0.25. The minimum mass
required to the placed on 20 kg mass to prevent
it from moving will be




freinds please solve this with explanation in copy​

Answer 1

◆ 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.