Question

A person pushing a horizontal, uniformly-loaded 29.05 kg wheelbarrow is attempting to get it over a step of height 0.410R, where R is the wheel\'s radius. What is the horizontal component of the minimum force P necessary to push the wheelbarrow over the step?

Answer 1

For the wheel barrow to go above the step the vertical component of the reaction on the wheel from the step must be greater than the weight of the wheel barrow and we also know that the reaction force at an edge acts along the normal to the contact surface (with the step) so the reaction force will be acting towards the center of the wheel

From the below image it can be seen that (theta = A)

sin A = 0.59,cos A = 0.807

we know that N sinA =29.05*9.8 =284.69N

N = 284.69/sin A = 284.69/0.59 = 482.525N

horizontal force = N cosA=482.525*0.807=389.398N

So the required force is 389.398N