Question

A container of mass 200 kg rests on the back of an open truck. If the truck accelerates at 1.5 m/s², what is the minimum coefficient of static friction between the container and the bed of the truck required to prevent the container from sliding off the back of the truck?

Answer 1

Hii dear,

# Answer- μ = 0.153

# Explaination-
# Given-
Mass of the truck, m = 200 kg
Acceleration, a = 1.5 m/s^2
Accn due to gravity g = 9.8 m/s^2

# Solution-
Normal reaction here,
N = mg
Frictional force,
F = ma

For container to stay on the truck without sliding,
F <= μN

Minimum coefficient of friction is given by,
F = μN
ma = μmg
μ = a/g
μ = 1.5/9.8
μ = 0.153

Minimum coefficient of friction is 0.153.

Hope that helps you.