John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a bri

Question

John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of ? = 15.5° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 420 N is exerted at the center of the wheel, which has a radius of 19.5 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive xaxis to be pointing to the right.

a) What force must John apply along the handles to just start the wheel over the brick

b) What is the force (magnitude and direction) that the brick exerts on the wheel just as the wheel begins to lift over the brick

Explanation / Answer

Just as the wheel begins to leave the ground, the normal force under it is zero. The brick at that point is resisting the downward weight of Rachel and the wheelbarrow as well as the downward and leftward (in my diagram) components of the applied force F.

The vertical distance from the corner of the brick to the center of the wheel is
h = 19.5cm - 8cm =11.5 cm,
and the horizontal distance from the corner of the brick to the center of the wheel is
d = (19.5² - 11.5²) cm = 15.74 cm

Sum the moments about the corner of the brick. At equilibrium, the clockwise moments must equal the counterclockwise moments:
(420N + Fsin15.5º) * 15.74cm = Fcos15.5º * 11.5cm
which solves to
F = 960.87 N     (a)

(b) R = ( (420 + 960.87sin15.5º)² + (960.87cos15.5)² ) N
R = 1146.89 N 1147 N magnitude
= arctan( (420 + 960.87sin15.5º)/(960.87cos15.5) )

= arctan(0.730)

= 36.129º above horizontal

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