Static Equilibrium rope and ball problem

AI Thread Summary
The discussion revolves around solving a static equilibrium problem involving a ball suspended by two ropes. The ball's weight is 11 kg, with rope A making a 22-degree angle to the vertical and rope B at an unspecified angle. Participants emphasize the importance of creating a free body diagram and applying equilibrium equations to find the tensions in both ropes. The correct approach involves breaking down the tensions into their x and y components and setting up simultaneous equations to solve for the tensions. Ultimately, the calculated tensions are approximately 166.9 N for rope A and 78.12 N for rope B.
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[URGENT]Static Equilibrium problem

A 11kg ball is supported from the ceiling by rope A. Rope B pulls downward and to the side on the ball. If the angle of A to the vertical is 22 degree and if B makes an angle of degree to the vertical find the tension in rope A.



Homework Equations


sum Force = 0
Fy= T sin theta-mg = 0
sum of Torque = 0


The Attempt at a Solution


I have no idea on how to solve this at all!
one thing I know is that the tension in Rope A is bigger than Rope B, maybe.

Thanks.
 

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Make a free body diagram. That means you draw a circle on a piece of paper and consider all the forces acting on the ball. You have the tension pulling the ball up, the weight (mass times gravity) pulling down, and the other tension pulling the ball to the side.

Next, make an important observation: the ball is not moving.

This means that the sum of all the forces acting on the ball must equal zero. So that's an equation.

T_a + T_b + mg = 0

But actually, this is two equations because these are vectors...

x: T_{a,x} + T_{b,x} + mg_x = 0

y: T_{a,y} + T_{b,y} + mg_y = 0

You know mg and you can calculate the two tensions because you are given the angles. You need to break the tensions into the x and y components of tension (by using Sin, Cos and the given angles). Then it's easy to solve for the x and y components of the tension in rope A.

Once you have the components of tension, vector-add them to find the total tension in rope A.

Make sense?
 


Ok, let's see:
for A:
x=mg cos theta + mg
y=mg cos theta + mg

answer = square root x^2+y^2 ?

Sorry, I'm a bit lost.
 


First of all, you can't use Cos for both the x and y components of tension. Maybe that was a typo?

Also, you are ignoring the tension from rope B in your equations. You won't be able to solve this unless you include them. So rewrite your equations to consider the x components of A, B, and gravity; then the y components. Then set these equations both equal to zero (it's zero because the ball is not moving, therefore it is at equilibrium therefore the sum of the forces must be zero). Then you just solve the system of equations to find your x and y components of tension for A.

You don't use the T_a = \sqrt{T_a,x^2 + T_a,y^2} part until you have solve for x and y.
 


Ta cos 22 + Tb cos 53 + 11*9.8 = 0
Ta sin 22 + Tb sin 53 = 0
solve simultaneously and
I got the answer!
Ta=166.9
Tb=78.12

thank u so much!
 
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