A student applies a force of 150 N on a 15 kg box by using a rope held at an angle of 35º above the horizontal. The coefficient of friction between the box and the surface is 0.90...

Question

A student applies a force of 150 N on a 15 kg box by using a rope held at an angle of 35º above the horizontal. The coefficient of friction between the box and the surface is 0.90.

a) Calculate the vertical and horizontal components of the 150 N force.

b) Calculate the acceleration of the box.

Answer 1

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Answer 2

a) Components of applied force F_p are F_px = 150 cos(35^o) = 123 N, F_py = 150 sin(35^o) = 86 N.

b) Sum the forces in both directions. Vertically, ΣF_y = F_N + 86 - mg = 0, F_N = 15*9.8 - 86 = 61 N.

Friction force f = μF_N = (0.9)(61) = 55 N. Horizontally, ΣF_x = 123 - f = ma, so 15*a = 123 - 55 = 68 N. Acceleration a = 4.5 m/s².