Question

A block is at rest on an inclined plane (shown in diagram). Coefficients of friction are μ-02, μ,-0.17. Gravity is 9.8 m/s2 Okg What is the normal force on the box? Let's assume friction were not a factor-the box would slide down the ramp. What is the magnitude of the force that causes the box to24 down the ramp? What is the static friction force on the box? Will the box slide without being pushed down the ramp? What is the magnitude of kinetic friction on the box? What is the acceleration of the box down the ramp with friction? 1. 2. 3. 4. 5.

Answer 1

Normal force N = 50 times 9.8 times cos 24 degree N = 447.64 N force cause the box to down the ramp = 50 times 9.8 times sin 24 degree = 199.3 N Static frictional force f_s = mu_s N = 0.2 times 447.64 f_s = 89.53 N Kinetic frictional force f_k = mu_k N = 0.17 times 447.64 f_k = 76.1 N sigma F_x = ma 199.3 - f_k = 50 times a a = 199.3 - 76.1/50 = 2.464 m/s^2