Question

A person pushing a uniformly‑loaded 34.7kg wheelbarrow of length L with pushing force P is attempting to get it over a step. The maximum horizontal force that the person can apply is P=617N. What is the maximum height h of the step, as a fraction n of the wheel's radius R , that the person can get the wheelbarrow over?

Answer 1

Draw a line from the centre of the wheel to the tip of the step, where the wheel touches the step...Let the angle between the vertical axis of wheel..and the line you have drawn...be {theta}

so relation will be Now, R ( 1 - cos {theta} ) = n R in order to find the value of n we need to find theta

Now, for the wheel to topple...the the force F, must be able to balance the counter clockwise torque of the WEIGHT (active vertically downwards on wheel)....about the point of contact of wheel and step....
So....F x R cos {theta} = (W/2) x R sin {theta}
F = W tan {theta} = ( 34.7 * 9.81 ) tan {theta} = 617 N

tan {theta} = 1.81438

theta = 61.1386

so now, n =  ( 1 - cos {theta} ) = 1 - cos 61.1386 = 0.5173

so maximum height = 0.5173 R