Question

A $30kg$ mass is initially at rest on the floor of a truck. The coefficient of static friction between the mass and the floor of truck in $0.3$ and coefficient of kinetic friction is $0.2$. Initially the truck is travelling due east at constant speed. Find the magnitude and direction of the friction force acting on the mass, if :
(Take $g=10m/s^2$). The truck accelerates at $1.8m/s^2$ eastward:

• A. $54N$ (due east)
• B. $90N$ (due east)
• C. $54N$ (due west)
• D. $60N$ (due east)

Answer 1

Answer: (A)

$54N$ (due east)

Coefficient of static friction ${\mu }_s=0.3$

static friction force

$f_s\le 0.3\times 10\times 30f_s\le 90N$

this means that mass will not move until the pseudo force overcomes the static friction

to move the block pseudo force must be greater than 90 N

lets calculate the Pseudo force

$F_p=-ma=30\times 1.8=54N$

$F_p<f_{s}max$

Friction force is self adjustable

friction force = pseudo force = 54 N

opposite to pseudo force means in the direction of motion (due east)

Option A is correct.