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N_L_
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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
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