# Three blocks suspended on 2 pulleys to form a M shape

## Homework Statement

A block with a mass of 25 kg is hung midway between two pulleys, with a rope connecting a hook at the top of the block to each of the two pulleys. On the other side of the pulleys, both ropes are connected to blocks with masses of 15 kg. The 25 kg mass in the middle sags downward until the 3 blocks and 2 pulleys form the shape of an “M”. If the pulleys are 45 cm apart, how far below the level of the pulleys does the top of the 25 kg mass need to be in order for the system to be in equilibrium?

Sum of Forces =0

## The Attempt at a Solution

I have calculated the tensions in one rope to be 147N using 15kg*9.8m/sec^2 and the Fg of the middle block is 245N I know I need to find the angles of the rope to the ceiling by splitting the tension of the rope in x and y directions, but I can not figure out how to get those values.

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The magnitude of vectors A and B:
A=25g N
B&C=15g N

$\vec{A}=\vec{B}+\vec{C}$

From above equation you can find the direction of B and C vectors for equilibrium.

If I come up with the forces add them 147N+147N≠245N So What am I doing Wrong?

Yes, it's wrong.
You have to think in vectors, which has magnitude and direction.
If the vectors are in same direction, ordinary arithmetic can be performed on them.

Another point is that the vector can be resolved/replaced by its components.
The best directions of vector components is in X(horizontal) and Y(vertical)direction.

I know that vectors are comprised of a magnitude and direction or can be written in component form. I just do not know the direction of the tension on the middle block so I do not know how to put it in component form without direction. What I am haing the most trouble with is how to separate that tension from the rope onto the middle block which is 147N into its' components. Am I trying to figure this out the wrong way? Could you please tell me which direction the tension is in or how to figure the direction out?

Let F be the tension.
There are 2 components that you can do normal arithmetics on identical direction.
Please note that there is another tensin on mirror side.
The weight vector is in Y direction.

http://imageshack.us/a/img805/2684/20946647.jpg [Broken][/QUOTE]

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Is weight vector is in the negative y direction? I have drawn all this out, but I do not have angle theta. Theta is not given in the problem, thats something I need to figure out. If the weight vector is in the negative direction it would have an equal and opposite force in the positive direction which would be 245 N correct? But you cannot have a triangle with side larger than its' hypotenuse so would force in the positive y direction half of 245N since it is beeing supported by two ropes?

If the force in the positive y direction is half, then the angle would equal 56.4 degrees and the height would be 22.5*tan(theta)= 33.9cm from the ceiling. Is that right?

To make the body in equilibrium, the sum of vectors with negative Y direction is equal to sum of vectors in positive Y direction.

Likewise in X direction.

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so is this right <-245j>= 147(xi+yj)+147(-xi+yj) ? Now how am I supposed to solve for the components?

the tangent is to solve for the height of the block to the ceiling and the distance between the pulleys is 45cm so once I have theta I can solve for the distance using trig and half that distance

What is the vertical component of the tension of one of the cables?

is it half of the force from the middle weight

rl.bhat
Homework Helper
the tangent is to solve for the height of the block to the ceiling and the distance between the pulleys is 45cm so once I have theta I can solve for the distance using trig and half that distance
You can find θ using the parallelogram law of Forces.
R2 = P2 + Q2 + 2PQcosθ.

There are 2 components forces upward each of 15g N tension.
Total positive upward force is - 2 x 15g x Sinθ
Downward force = 25g

You can get θ.
From this you can get height which equal to L/2(Tanθ)