# Physics: an elephant pushing a shopping cart

1. Feb 18, 2006

### N_L_

My question is whether the FBDs I have for the following problem are correct or not:

A 320 kg homeless elephant pushes a 700 kg shopping cart. The coefficient of static friction between the feet of the elephant and the ground is 0.98. The elephant (and the cart) accelerate at 0.27 m/s^2. Assume the elephant is pushing hard (just before its feet start slipping).

In the figure the cart is on the left and the elephant is on the right.

In the FBD for the elephant I have Fg pointing down, Fn pointing up, the force of the elephant on the cart pointing left, the force of the cart on the elephant pointing right, and the force of friction pointing right.

In the FBD for the cart I have Fg pointing down, Fn pointing up, the force of the cart on the elephant pointing right, the force of the elephant on the cart pointing left, and the force of friction pointing right.

Correct? Incorrect? Not even close

2. Feb 18, 2006

### d_leet

Pretty close but the forces that the elephant causes should not be in the fbd for the elephant and it is the same way for the cart. Only put the forces acting on the objects.

3. Feb 19, 2006

### N_L_

Okay. Thank you for the clarification.

4. Feb 19, 2006

### lightgrav

Only include the Forces which act on the object (singular) that it is a FBD of.

5. Feb 23, 2006

### N_L_

Is this correct?

A 320 kg homeless elephant pushes a 700 kg shopping cart. The coefficient of static friction between the feet of the elephant and the ground is 0.98. The elephant (and the cart) accelerate at 0.27 m/s^2. Assume the elephant is pushing hard (just before its feet start slipping).

In the figure the cart is on the left and the elephant is on the right.

a) The force of the cart on the elephant.
b) The force of the elephant on the cart.
c) The force of friction on the cart.
d) The coefficient of friction for the cart.
e) Is this static or kinetic friction?

Given:

Elephant = 320 kg
Cart = 700 kg
Mus = 0.98
acceleration = 0.27 m/s^2

What I got:

Fapplied = the force of the elephant on the cart
Fc/e = the force of the cart on the elephant
Ffr = force friction

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

Ffr = 0.98*m*g = 3073.28 newtons

Fapp - 3083.28 = 320 kg * 0.27

Fapp = 3159.68 newtons

Fc/e - Ffr = m*a
3159.68 - Mu (9.8 * 700) = 700 * .27
Mu = .433

a. 3159 newtons

b. The force of the cart on the elephant is equal and opposite to that of the elephant on the cart.

c. Ffr = 3073 newtons

d. .433

e. It has to be static friction in order for the wheels of the cart to turn?