Find the tension in each cable necessary to keep the block from moving (either sideways or downward)

Question

A 900-lb block is being suspended and held stationary by two steel cables as shown.  The first cable (to the left of the block) makes a 70 degree angle with the horizontal, while the second cable (to the right of the block) makes a 65 degree angle with the horizontal.  At point B (where the weight is being held), there are three forces acting: R, S, and W.  Find the tension in each cable necessary to keep the block from moving (either sideways or downward).

 

Please try to explain the steps.

I know so far, Area=1/2absin(angle c)

(8)(6)(1/2(x)(x)sin60)=1590

Answer 1

First, we need to take a step away from the area idea - that is not what we ned to use here.

 

What we should do is to take a look at the forces in each direction (like you would do with projectile motion)

 

If we knew what the tensions were, what would we do to find the components? Let's say the the tension on the left is TL and the tension on the right is TR.

 

The left tension points up and left and the right tension points up and right. Kind of like

 

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We want to find the components of each tension. We assume that "to the right" and "up" are positive.

 

The x component of TR is TR * cos(65) (from Soh Cah Toa)

The x component of  TL is -TL * cos(70) for the same reason. It is negative because positive forces point to the right

 

These are our only x direction forces. So now we should look in the y direction.

 

TRy = TR * sin(65)

TLy = TL * sin(70)

 

We also have the weight of the block, which points downwards, so that is -900 lb

 

Now we write out a sum of forces (or net force) in the y direction.

 

ΣFy = TR * sin(65) + TL * sin(70) - 900 = 0  (this is equal to 0 because everything is stationary)

 

We have 1 equation and 2 unknowns, TR and TL, so we need another equation. Let's look in the x-direction now and do another sum of forces

 

ΣFx = TR * cos(65) - TL * cos(70) = 0

 

Once again, the sum of forces is equal to 0 because the block is not moving.

 

Now we have 2 equations and 2 unknowns. How do we deal with that? The idea is to solve for one unknown in 1 equation, then plug it into the other. 

 

From the second equation, 'solving' for TR, we get

 

TR = TL * cos(70) / cos(65)

 

Now substitute this into the other equation for TR. We now have 1 equation and 1 unknown

 

[ TL*cos(70)/cos(65) ] * sin(65) + TL * sin(70) - 900 = 0

 

add 900 to both sides

 

[ TL * cos(70) / cos(65) ] * sin(65) + TL * sin(70) = 900

 

We can solve for TL. To do this, we factor out TL on the left

 

TL * [ cos(70) / cos(65) * sin(65) + sin(70) ] = 900

 

Then divide by that big bunch of sines and cosines

 

TL = 900 / [ cos(70) / cos(65) * sin(65) + sin(70) ]

 

We've now solved for TL. To find TR, we plug what we get for TL back into the previous equation

 

TR = TL * cos(70) / cos(65)

 

I hope this is clear! Please comment if you have any questions. Good luck!