How Much Force Does Stephen Need to Overcome the Brick with a Wheelbarrow?

[ramin86]

6. [PSE6 12.P.022.] Stephen is pushing his sister Joyce in a wheelbarrow when it is stopped by a brick 7.83 cm high (Fig. P12.22). The handles make an angle of 15.0° below the horizontal. A downward force of 389 N is exerted on the wheel, which has a radius of 20.3 cm.

There was a similar problem in the book, so I used the formulas that they had in there.

(a) What force must Stephen apply along the handles in order to just start the wheel over the brick?
N
For this one, I did F= mg * sqrt(2rh-h^2)/(2r-h), with mg = 389N, and the radius of .203m, and a height of .0783m, and I came up with 190N, however, the answer turned out wrong.

(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? Assume in both parts that the brick remains fixed and does not slide along the ground.
Magnitude
kN
For this, I did R = sqrt((mg)^2) + F^2), again turned out wrong

(c)Direction
And for this, I used tan = mg/f and solved for theta, however, it turned out wrong as well.
° above the horizontal (to the left)

If someone can please help, it would be greatly appreciated. Here is a link to the diagram: http://www.webassign.net/pse/p12-11.gif

 

[wais]

To help with this problem, let's break it down into smaller parts and use some basic principles of static equilibrium. First, let's draw a free-body diagram for the wheel and the brick. The forces acting on the wheel are the downward force of 389 N, the normal force from the ground, and the force from Stephen pushing on the handles. The forces acting on the brick are the normal force from the ground and the force from the wheel pushing on it.

Now, let's analyze the forces in the vertical direction. Since the wheel is not moving vertically, the sum of the vertical forces must be equal to zero. This means that the normal force from the ground must be equal to the downward force of 389 N.

Next, let's look at the forces in the horizontal direction. Since the wheel is not moving horizontally, the sum of the horizontal forces must also be equal to zero. This means that the force from Stephen pushing on the handles must be equal in magnitude and opposite in direction to the force from the wheel pushing on the brick.

Now, let's look at the forces at the point where the wheel and the brick meet. At this point, the forces from the wheel and the brick must cancel each other out in order for the wheel to be in static equilibrium. This means that the force from the wheel must be equal in magnitude and opposite in direction to the force from the brick.

Using these principles, we can now solve for the force that Stephen must apply along the handles in order to just start the wheel over the brick. Since the force from the wheel and the brick must cancel each other out, we can set the magnitude of these forces equal to each other. This means that F = mg * sqrt(2rh-h^2)/(2r-h) = 389 N. Solving for F, we get F = 190 N.

For part (b), we can use the same principles to solve for the force that the brick exerts on the wheel. Since the forces from the wheel and the brick must cancel each other out, the magnitude of the force from the brick must be equal to the magnitude of the force from the wheel, which we found to be 190 N. The direction of this force will be opposite to the direction of the force from the wheel, which is 15.0° below the horizontal. This means that the force from the brick will be 15.0° above the horizontal