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

The acceleration of the body is 7.0105

Explanation:

From the free body diagram of the given case

Vertical component of the applied force , = F cos(30°) = = 10

Horizontal component of the applied force , = F sin(30°)

Balancing the normal force(N) from the wall and in horizontal direction.

N = = F sin(30°) = = 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 - mg= ma

∴ 10 - 0.5 - (1×9.81) = 1×a

∴ a = 7.0105

the answer would be 32*f

snell's law

refractve index = sin incidence / sin refraction