physics mod2top5

Question

A person pushes a 14.0kg lawnmower at a constant speed with a force of 88.0N directed along the handle, which is at an angle of 45° to the horizontal. Draw a free body diagram and calculate the friction force on the mower. Use the summation of the forces in the x-direction and the y-direction to find your answer.

Answer 1

For the free body diagram, you have the normal force acting straight up, the weight of the mower acting straight down, the force of the push acting diagonally (which will have an x and y component), and the friction force acting backwards. The friction force = F_friction = μ_k * F_N. Since we don't know the coefficient of friction, we'll need to sum the forces to solve. Since the mower is moving at a constant speed, we know that the acceleration is 0 so the sum of forces in the x and y direction will equal 0.

ΣF_y = 0 = F_N - W - F_pushsin45° = F_N - mg - F_pushsin45° → F_N = mg + F_pushsin45° = 14.0kg(9.8 m/s²) + 88.0N (sin45°) = 199.425

ΣF_x = 0 = F_pushcos45° - F_friction → F_friction = F_pushcos45° = 88.0N (cos45°) = 62.225N