Question

A 1.50-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction between the block and the wall is 0.54. What is the minimum compression in the spring to prevent the block from slipping down?

Answer 1

Answer:

0.032 m

Explanation:

Consider the forces acting on the block

m = mass of the block = 1.50 kg

$f_{s}$ = Static frictional force

$F_{n}$ = Normal force on the block from the wall

$F_{s}$ = Spring force due to compression of spring

$F_{g}$ = Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N

k = spring constant = 860 N/m

μ = Coefficient of static friction between the block and wall = 0.54

x = compression of the spring

Spring force is given as

$F_{s}$ = kx

From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as  

$F_{n}$ = $F_{s}$

$F_{n}$ = kx                                             eq-1

Static frictional force is given as

$f_{s}$ = μ $F_{n}$

Using eq-1

$f_{s}$ = μ k x                                                eq-2

From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as

$f_{s}$ = $F_{g}$

Using eq-2

μ k x = 14.7

(0.54) (860) x = 14.7

x = 0.032 m