Question

A1 kg-body moves upwards across the vertical wall due to the force F= 20 N applied at the angle a30° with vertical. Find the acceleration of the body if the coefficient of friction is k 0.05

Answer 1

Answer:

The acceleration of the body is 7.0105 $\frac{m}{s^{2} }$

Explanation:

From the free body diagram of the given case

Vertical component of the applied force , $F_{v}$ = F cos(30°) = $\dfrac{\sqrt{3}F }{2}$ = 10$\sqrt{3}$

Horizontal component of the applied force , $F_{h}$ = F sin(30°)

Balancing the normal force(N) from the wall and $F_{h}$ in horizontal direction.

N = $F_{h}$ = F sin(30°) = $\dfrac{F}{2}$ = 10 N

∵ The body is accelerating in vertical direction , the friction force acting on the body should be maximum.

∴ Maximum frictional force , f = k×N = 0.05×10 = 0.5 N

Let 'm' be the mass of the body and 'a' be the acceleration of the body in vertical direction and 'g' be the acceleration due to gravity.

∵ Body is accelerating in vertical direction , By Newton's 2nd law

$F_{v}$ - f - mg= ma

∴ 10$\sqrt{3}$ - 0.5 - (1×9.81) = 1×a

∴ a = 7.0105 $\frac{m}{s^{2} }$