find the tension

Question

hree masses are connected by strings as shown in the attached diagram. Their masses are M1 = 1 kg, M2 = 2 kg, and M3 = 3 kg. The coefficient of friction between the blocks and the surface is 0.25. If the blocks are pulled with a constant force and move at a constant velocity if 1 m/s, what are the tensions in the strings?  

Answer 1

Alright, Matt, I'm going to take a crack at this without the attached diagram. I think the text tells us everything we need to know, even without the diagram. Here are my assumptions:

1. every block is touching the surface and they're in the order 3, 1, 2 from left to right
2. block 1 is attached to block 2, and block 2 is attached to block 3
3. The string to pull all three blocks is attached to block 2 and pulls from the right-hand side
4. All of the strings are horizontal

To find the tension on the right-hand string on block 3, you find the total force acting on the system to the left, which is the frictional force on all of the blocks. The system moves at a constant speed, so it's in equilibrium; this means that

-Ff(2) - Ff(3) - Ff(1) + Ft(right side of block 2) =0

or

F(right side of block 2) = Ff(2)+Ff(3) + Ff(1)

Ff for each block can be determined using Ff = (mu)Fn. Since it's a horizontal surface, Fn=mg. We can now find this for each block:

Ff(2) = (0.25)(2 kg)(9.8 m/s^2) = 4.9 N

Ff(1) = (0.25)(1 kg)(9.8 m/s^2) = 2.45 N

Ff(3) = (0.25)(3 kg)(9.8 m/s^2) = 7.35 N

F(right-hand side of 2) = 4.9 N + 2.45 N + 7.35 N = 14.7 N

The net force equation for the rope connecting blocks 1 and 2 (the rope furthest to the left in my picture) would be

-Ff(2) + FT(blocks 1 and 2) = 0

FT(blocks 1 and 2) = Ff(2) = 9.8 N

For the final rope, the one connecting blocks 2 and 3 (the "middle" rope), the net force equation would be

-Ft(blocks 1 and 2) - Ff(1) + FT(blocks 2 and 3) = 0

- 9.8 N - 2.45 N + FT(blocks 2 and 3) = 0

FT(blocks 2 and 3) = 7.35 N

This could also be done with ratios: the rope between 1 and 2, for example, only needs to pull 2/6 of the total force, so the tension is 2/6 x 14.7 or 9.8 Newtons.

Good luck!

A