# Forces problem: a donkey pulling a cart

1. Feb 23, 2006

### N_L_

I worked out the following problem, but I am unsure of my answers. Do they look correct?

Part A

A 255 kg donkey is pulling on a cart with a horizontal force of 1940 N. The mass of the cart is 520 kg and the cart is rolling at a constant velocity of 0.72 meters / second to the left. Find:

a) The coefficient of kinetic friction for the wheel's bearings on the cart.
b) The force exerted on the donkey by the surface.

What I have:

constant velocity = zero acceleration

The sum of the forces on the in the X direction = Fapp - Ffr = 0

1940 N - Mu (9.8 * 520) = 0

Mu for the cart = .381

The force exerted on the donkey by the surface = Fnormal. Fn = m*g = 2499 N

Part B

The same donkey is pulling on the same cart, but now the cart is accelerating at 0.24 m/s^2 to the left. Find:

a) The force by the donkey on the cart

b) The force by the surface on the donkey

The sum of the forces in the X direction = Fapp - Ffr = 255 * 0.24

Fapp - (Mu)(9.8)(255) = 61.2

From the previous part of the problem, Mu = .776

Fapp = 2000.42 N

Fn = 9.8 * 255 = 2499 N

2. Feb 24, 2006

### daniel_i_l

A)
The first part looks good, but in b the surface exerts force on the donkey on the X axis to. If it didn't then the donkey would just tread in place without being able to move.
B)
You wrote: Fapp - Ffr = 255 * 0.24 , but you're looking at the forces on the cart. Whose mass are you using?

3. Feb 24, 2006