Suspended object, two rope system

In summary: FIn summary, the magnitude of the force due to the ropes is irrelevant, but the angle at which the force is applied is important.
  • #1
Nal101
16
0

Homework Statement



1) What is Y-component of the force due to rope 2
2) What is the magnitude of F2
3) What are the X components of the force due to rope 1 and rope 2
hangingweight1.gif

hangingweight_force3.gif


mass = 60kg
[tex]\theta[/tex] = 20 degrees

Homework Equations


F = ma
ma = T-W


The Attempt at a Solution


1) mass * gravity = 60 * -9.8 = -588
To find the Y component, take -588 * sin(90-20) (Should the angle be off of horizontal or vertical, currently set to horizontal)

2) No clue, the actual term 'magnitude' seems to confuse me, so if someone could briefly define what the magnitude of a force is with relevance to the actual force, that would be great.

3) Same as 1, just substitute sin with cos (for rope 2)
For rope 1, is it 0? That doesn't seem right though, because the rope offset the block to create the angle of 20 degrees.
 
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  • #2
It would be easier to see what is going on if you resolved F2 into vertical and horizontal components. Then your diagram will look like a cross with four arms at right angles. If the net force is zero on the block, this means that the horizontal and vertical arms are

1. Irrelevant to the problem.
2. Equal in size and opposite in direction, in pairs
3. Don't know.

What do you think?
 
  • #3
Obviously, if you assign the entire picture as the system, the net force is going to be zero. What I am trying to figure out is when the system is defined as just the block, and to find the separate external forces as noted in the questions. It's the 'arms' separate force that I need to calculate, knowing the basic rules of equal & opposite (and therefore cancellation) is true, but doesn't help much if at all on the calculations.. other than the fact that on question 3, the numbers should be the same, but one is negative and one is positive.
 
  • #4
Actually the "equal and opposite forces" is probably the key to solving this problem.

Understanding that the block is currently stationary, then what can you conclude about the net forces on the block?

After breaking F2 into vertical and horizontal components you can easily observe how those forces interact with F1 and gravity.

I'm not sure what you're asking about the relativity of magnitude of a force to the actual force. As far as I'm concerned that is nearly the same thing, the only difference being that since force is a vector, an "actual force" would usually be accompanied by a direction, usually an angle. Your question only cares about the magnitude, or the "scalar part". What is that tension in the rope that is keeping the block stationary?
 
  • #5
Question: Are my calculations correct? If not, where did I go wrong?

Understanding that the block is currently stationary, then what can you conclude about the net forces on the block?
I already said that they cancel each other out. i.e. the net force acting on the block is zero.

After breaking F2 into vertical and horizontal components you can easily observe how those forces interact with F1 and gravity.
I know, my question is HOW.
 

FAQ: Suspended object, two rope system

How does a suspended object in a two rope system maintain its stability?

In a two rope system, the suspended object maintains its stability through the principle of equilibrium. The two ropes exert equal and opposite forces on the object, keeping it in place and preventing it from falling.

What factors affect the tension in the ropes of a suspended object in a two rope system?

The tension in the ropes of a suspended object in a two rope system is affected by the weight of the object, the angle at which the ropes are attached to the object, and the distance between the two attachment points on the object.

How does the angle of the ropes affect the stability of a suspended object in a two rope system?

The stability of a suspended object in a two rope system is directly affected by the angle of the ropes. The closer the ropes are to being parallel, the greater the stability of the object. If the angle becomes too small, there is a risk of the object slipping off one of the ropes.

Can the weight of the object in a two rope system be greater than the tension in the ropes?

No, the weight of the object cannot be greater than the tension in the ropes in a two rope system. If the weight of the object exceeds the tension in the ropes, the system will become unbalanced and the object will fall.

What are the practical applications of a suspended object in a two rope system?

A suspended object in a two rope system has various practical applications, such as in lifting and rigging operations, rock climbing, and construction of suspension bridges. The stability and equilibrium of the system make it a useful tool for safely moving and supporting heavy objects.

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