Forces problem: a donkey pulling a cart

[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

 

[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?

 

[N_L_]

Thank you for your help.

 

[jimbat]

This is an ancient problem that I first read as a teenager in the 50s in "The World of Mathematics" by Newman. It is not an ordinary physics problem involving friction, etc. The conundrum, properly stated, is that the force the the donkey exerts on the cart is equal and opposite to the force that the cart exerts on the donkey. These forces add to zero according the the Third Law, so how can the donkey pull the cart? At parties in grad school in the 60s, not one physics professor was able to answer this problem correctly in terms of Newton's Laws. The problem is the indirection of referring to the 3rd law. It is the 2nd law that controls the motion of the cart. The motion of a body is controlled by all the forces exerted on *it*, not the forces it exerts on other bodies. When I was a professor, not one student ever got this problem completely right even though they were asked it on every quiz, and tutored in section. All I had to do was replace the donkey pulling a cart with a bee landing on a flower, or a pitcher throwing a ball to completely confuse them, because they didn't understand Newton's laws. Few do. Almost all of us are Artistotelians in daily life. The fact that the reciprocal forces add to zero is of course just a restatement of the law of conservation of momentum.

 

[rwisz]

Your calculations for Part A seem to be correct. The coefficient of kinetic friction for the wheel's bearings on the cart is indeed 0.381, and the force exerted on the donkey by the surface is 2499 N.

For Part B, your calculations for the force by the donkey on the cart and the force by the surface on the donkey appear to be correct as well. However, it would be helpful to label the units for the force in the X direction as N (newtons) rather than just stating "255 * 0.24". Additionally, it would be beneficial to show the steps of how you arrived at the values for Mu and Fapp, as these are important components in understanding the problem. Overall, your answers seem to be correct, but it is always a good idea to show all of your work and clearly label your units for clarity and accuracy.