Physics. How do you work out the tension in a rope between two objects with one object being pulled.?

Question

Two carts connected by a light, inextensible rope are being accelerated across a tarmac by a tractor. The leading cart has a mass of 35kg; the trailing cart has a mass of 25kg.The coefficient of friction on the tarmac is 0.20. The tractor exerts a pulling force of 190 N on the cart of 35kg. The breaking tension in the rope is 150 N.

a. Draw a force vector diagram for each of the two masses.

b. Calculate the magnitude of the acceleration of the two carts.

c. Calculate the tension in the rope which connects the two masses.

I have worked out the acceleration to be 1.21 m/s^2

I don't know how to work out question c.

Can anyone help?

Answer 1

a. For your force vector diagram, draw two rectangles to represent the carts and connect then with a line to represent the rope connecting the two carts. The first rectangle represents the 35 kg cart and the second one reresents the 25 kg cart. Represent the total pulling force as an arrow connected at the front of the 35 kg cart going to the right. Label thisas P. Represent the friction force on the 35 kg cart as a half arrow under the cart which is pointing to the left.  Label this as F1. Represent the friction force on the 25kg cart as a half arrow under the cart pointing to the left.Label this as F1. Now, represent the inertial force of the 35 kg cart as an arrow above the cart which is pointing to the left. Label this as F3. Represent the inertial  force of the 25 kg cart as an arrow above the cart point to the left. Label this as F4. Represent the tension on the rope between the carts as an arrow above the connecting line pointing to the right. Label this as T.

b. To calculate the acceleration we must first calculate the forces.

Let:

a = the acceleration

g = acceleration og gravity = 9.81 m/sec^2

M1 = mass of the first cart = 35 kg

M2 = mass of the second cart = 25 kg

f = coefficient of frction = 0.2

F1 = f M1 g = 0.2 x35 x 9.81 = 68.67 N

F2 =  f M2 g = 0.2 x 25 x 9.81  = 49.05 N

F3 = M1 a = 35a

F4  = M2 a =  25a

P = 190 N

P = F1 + F2 + F3 + F4 = 68.67 + 49.05 + 35a + 25a

190 = 117.72 + 60a

a = (190 -117.72)/60 = 1.205 m/sec^2

c. For the tension on the line between the two carts

T = F2 + F4 = 49.05 + 25x1.205 = 79.175 N

Answer 2

Forces on 1st cart (x direction only):

(190N-T-friction)=(190-T-0.2*35*g)

Newton's first law:

(190-T-0.2*35*g)/35=a (which you have found)

Forces on 2nd cart (x dir.):

(T-friction)=T-0.2*25*g

(T-0.2*25*g)/25=a

So you have:

(190-T-0.2*35*g)/35=(T-0.2*25*g)/25

An equation with one variable which you should easily solve.