Physics: an elephant pushing a shopping cart

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The discussion centers on the free body diagrams (FBDs) of a 320 kg elephant pushing a 700 kg shopping cart, focusing on the forces acting on each object. The initial FBDs presented were incorrect because they included forces that the elephant and cart exert on each other, which should not be included in their respective diagrams. The correct approach is to only depict the forces acting on the object in question. Calculations reveal that the applied force by the elephant is approximately 3159.68 Newtons, and the friction force is 3073 Newtons, indicating that static friction is in effect as the cart's wheels must turn. The final answers regarding forces and friction coefficients were confirmed to be correct, emphasizing the importance of accurately representing forces in FBDs.
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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 :confused:
 
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N_L_ said:
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 :confused:


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.
 
Okay. Thank you for the clarification.
 
Only include the Forces which act on the object (singular) that it is a FBD of.
 
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?



Are these answers correct?
 
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