Here is the problem: Stephen is pushing his sister Joyce in a wheelbarrow when i
ID: 1746080 • Letter: H
Question
Here is the problem:Stephen is pushing his sister Joyce in a wheelbarrow when it isstopped by a brick 80 cm high (Fig. P12.18). The wheelbarrowhandles make an angle of 15.0 degrees below the horizontal. Adownward force of 400 N is exerted on the wheel, which has a radiusof 20.0 cm. (a) What force must Stephen apply along the handles tojust start the wheel over the brick? (b) What is the force(magnitude and direction) that the brick exerts on the wheel justas the wheel begins to lift over the brick? In parts (a) and (b),assume the brick remains fixed and does not slide along theground.
Solutions?
Explanation / Answer
from the figure in the text book the forces are Fx = F cos15.0o Fy = Fsin15.0o as the wheel leaves the ground the ground exertsno force on it so we get Fx = 0 F cos15.0o - nx= 0 ........ (1) Fy = 0 F sin15.0o - 400 N +ny = 0 ........ (2) as we take the torques about the contact pointwith the brick so the needed distances will be b = R - 8.00 cm = (20.0 - 8.00) cm = 12.0 cm a = (R2 -b2) = ........ cm (a) as = 0 - Fx b + Fy a + (400 N) a =0 F = ........ N (b) using the equations (1) and (2) we get nx = F cos15.0o = ........ N amd ny = (400 N)+ F sin15.0o = ........ N n = (nx2 +ny2) = ........ N = tan-1(ny /nx) = ........o towardsthe left and upward Fy = Fsin15.0o as the wheel leaves the ground the ground exertsno force on it so we get Fx = 0 F cos15.0o - nx= 0 ........ (1) Fy = 0 F sin15.0o - 400 N +ny = 0 ........ (2) as we take the torques about the contact pointwith the brick so the needed distances will be b = R - 8.00 cm = (20.0 - 8.00) cm = 12.0 cm a = (R2 -b2) = ........ cm (a) as = 0 - Fx b + Fy a + (400 N) a =0 F = ........ N (b) using the equations (1) and (2) we get nx = F cos15.0o = ........ N amd ny = (400 N)+ F sin15.0o = ........ N n = (nx2 +ny2) = ........ N = tan-1(ny /nx) = ........o towardsthe left and upward