How Does Torque Affect a Cart Overcoming a Step?

[a_narain]

Homework Statement

A garden cart loaded with firewood is being pushed horizontally when it encounters a step 8.0 cm high, as shown in the figure. The mass of the cart and its load is 64 kg, and the cart is balanced so that its center of mass is directly over the axle. The wheel diameter is 60 cm. You are pushing the cart horizontally by a handle.

Homework Equations

torque = rFsin(angle)
Newton's laws

The Attempt at a Solution

I tried to find the torque that occurs as a result of the step applying a force on wheel. I originally thought that the force the person was applying was equal to the force that the wheel applied to the step. Thus, since the step is 8 cm high, I thought that the torque due to the step was .30(F)sin(74.5). But I am having trouble figuring out if this is equal to zero or something else?

 

[LowlyPion]

Welcome to PF.

What is the question?

 

[nlightner]

First of all, it is important to note that the torque due to the step is not causing any rotation in this scenario, since the cart is balanced and the center of mass is directly over the axle. Therefore, the torque due to the step can be ignored in this case.

To calculate the torque that the person is applying to the cart, we need to consider the force they are applying and the distance from the point of rotation (the axle) to the point where the force is being applied (the handle). This distance is the radius of the wheel, which is half of the wheel's diameter. So, the torque applied by the person is:

torque = rFsin(angle)
= (0.3 m)(F)(sin 74.5)
= 0.3Fsin 74.5

Now, we also need to consider the torque due to the weight of the cart and its load. This torque is causing a clockwise rotation, while the torque applied by the person is causing a counterclockwise rotation. In order to keep the cart balanced and prevent it from tipping over, these two torques must cancel each other out. Therefore, we can set the torque due to the weight equal to the torque applied by the person, and solve for F:

0.3Fsin 74.5 = (0.3 m)(64 kg)(9.8 m/s^2)(sin 90)
F = (0.3 m)(64 kg)(9.8 m/s^2)(sin 90) / (0.3 sin 74.5)
F = 204.8 N

So, in order to keep the cart balanced and prevent it from tipping over, the person must apply a force of 204.8 N to the handle.