Solve Mass Suspension by Component Method

[yoyo16]

Homework Statement

A mass of 100 kg is suspended by two ropes that make an angle of 60 degrees to the horizontal. If a horizontal pull of 200N, in a plane perpendicular to the plane of the other forces is applied find the tensions in the ropes after equilibrium has occurred.

Homework Equations

Component Method

The Attempt at a Solution

T1cos60=T2cos60
T2=(T1cos60)/(cos60)

Fx=-T1cos60+T2cos60
Fy=T1sin60+T2sin60-Fg

Sub in T2 into equation
Fy=T1sin60+T2sin60-Fg
= (T1sin60)+((T1cos60)/(cos60))(sin60)
= T1sin60+T1(0.866)
=T1(sin60+0.866)=(100)/(9.8)
T1=T2=565.8

Can someone please tell me if I did this right?

 

[vela]

It doesn't look it. You solved the problem without the 200 N force.

 

[yoyo16]

What would the 200 N force be?

 

[vela]

The one in the problem statement:

If a horizontal pull of 200N, in a plane perpendicular to the plane of the other forces is applied find the tensions in the ropes after equilibrium has occurred.

 

[yoyo16]

Yes but how would I put that in the equation? When using the component method and drawing the diagram where would I place the 200N?

 

[vela]

"In the plane perpendicular to the plane of the other forces." If the tensions and weights are initially lying in the xy-plane, the horizontal 200-N force would point in the z direction.

 

[yoyo16]

Would the 200N be for both ropes? And if the 200N was for the z direction, how would you find the tension of the ropes? Wouldn't the 200N be the tension then?

 

[Chestermiller]

This is a bit of a complicated problem geometrically. Before you start worrying about the forces involved, you need to consider in detail the 3D geometry (after the 200 N force has been applied). When the 200 N force pushes horizontally on the mass, the plane that had originally contained the two ropes and the mass rotates out of the vertical by an angle of, say, θ. The two ropes now have components in all three spatial directions. The simplest thing to do is to resolve the unit vector pointing in the direction along each of the ropes into components in the x, y, and z directions (in terms of i, j, k, and the angle θ). You can then draw a free body diagram of the mass, and do force balances in the x, y, and z directions in terms of θ. This will allow you to solve for the tensions in the ropes and the angle θ. Figuring out the geometry and resolving the rope unit vectors into components is the complicated part of this problem.

Chet

 

[vela]

You can also reduce it to a two-dimensional problem in the plane of the ropes. That makes the geometry a lot simpler.

 

[Chestermiller]

You can also reduce it to a two-dimensional problem in the plane of the ropes. That makes the geometry a lot simpler.

Yes. Excellent suggestion. I like that much better.

Chet

 

[yoyo16]

I'm still not understanding the concept of the 200N. Is it not possible to solve it without taking account of the horizontal pull. By using the horizontal pull, how would you use the component method? Can someone please try explaining this to me, I'm so confused. :0

 

[Chestermiller]

Step 1 is drawing a figure looking edge-on on the plane containing the ropes, after the 200 N force is applied. Can you please show us what your figure looks like.

Chet

 

[vela]

I'm still not understanding the concept of the 200N. Is it not possible to solve it without taking account of the horizontal pull.

Of course not. The problem is asking you to find the tension in the ropes when the system is in equilibrium when the 200-N force is applied. You can't possibly get the right answer if you omit that force.

By using the horizontal pull, how would you use the component method? Can someone please try explaining this to me, I'm so confused. :0

It seems to me you might not even understand the basic configuration of the ropes, mass, and force. If you don't understand that, you can't possibly hope to even begin analyzing the situation. It would help if, as Chet suggested, you show us a sketch (or two) of what you think is going on.