Is My Assumption About Rope Tension Correct?

[stuc]

Homework Statement

Homework Equations

My assumption:-
if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 5N load on it.Is it my assumption is correct?

The Attempt at a Solution

Tension of the other rope = 5N (is it correct)?? base on my assumption [/B]

 

[haruspex]

My assumption:-
if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 5N load on it.Is it my assumption is correct?

I would assume they mean the tension at the instant after cutting the other string.

 

[phinds]

Have you studied force vectors? I think your assumption is half correct and your conclusion is wrong. What you are right about is the hanging vertically. All of your statements about force are wrong.

 

[phinds]

I would assume they mean the tension at the instant after cutting the other string.

Hm ... could be. I didn't interpret it that way but I guess that could be right. stuc, my comments are not based on that assumption.

 

[stuc]

I would assume they mean the tension at the instant after cutting the other string.

...the tension at the instant after cutting the other string

how to calculate it? it mean my assumption is wrong

i just check an answer..the answer is 5N...but how to get it

 

[haruspex]

.the answer is 5N.

Hmm.. that does not fit with either interpretation. But I remain of the view that it's the instantaneous value they're after.
First, can you figure out the weight of the mass?

 

[phinds]

You have not answered my question about whether or not you have studied force vectors.

 

[stuc]

Hmm.. that does not fit with either interpretation. But I remain of the view that it's the instantaneous value they're after.
First, can you figure out the weight of the mass?

there is no weight given.

 

[stuc]

You have not answered my question about whether or not you have studied force vectors.

learn but always confuse when try to solve it

 

[phinds]

there is no weight given.

What is the definition of a Newton?

 

[stuc]

What is the definition of a Newton?

1N = the force needed to accelerate 1kg of mass

 

[stuc]

Attached is the calculation..any one can verify my ans (4.33N) because the given ans in 5N.

Update:Wrong Arrow for Fy

 

[stuc]

Attached other option ans

 

[haruspex]

As I posted, the first step is to determine the weight of the mass. No, it's not given, but there is enough information to find it. To do this, just consider the initial set up with two ropes. Write the equation for balance of forces in the vertical direction.

 

[Chestermiller]

Attached is the calculation..any one can verify my ans (4.33N) because the given ans in 5N.

Update:Wrong Arrow for Fy

You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

Chet

 

[stuc]

As I posted, the first step is to determine the weight of the mass. No, it's not given, but there is enough information to find it. To do this, just consider the initial set up with two ropes. Write the equation for balance of forces in the vertical direction.

My step to determine the weight of the mass..is it ok? (weight of the mass = 7.071N)

 

[stuc]

You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

Chet

weight of the mass = 7.071N

 

[haruspex]

My step to determine the weight of the mass..is it ok?
View attachment 91455

No, that doesn't work. Pythagoras applies when two vectors at right angles are to be added.
What is the component of each tension in the vertical direction?

 

[stuc]

No, that doesn't work. Pythagoras applies when two vectors at right angles are to be added.
What is the component of each tension in the vertical direction?

..tq for the info..will be calculate

 

[haruspex]

..tq for the info..will be calculate

Hint: you already calculated it in post #12.

 

[Chestermiller]

My step to determine the weight of the mass..is it ok? (weight of the mass = 7.071N)
View attachment 91455

In this figure, the 5 N should be the hypotenuse, not one of the other sides. You're using the wrong triangle.

 

[stuc]

Hint: you already calculated it in post #12.

Base on the above diagram (arrow edited),

5N x sin 60 + 5N x sin 60 - W = 0
therefore W= 2(5N x sin 60)
= 8.66 N

 

[stuc]

You got 4.33 N for the upward component of force contributed by each wire. But there are two wires. So what is the weight of the object? If this weight has to be supported by only one wire, what would the tension in that wire have to be?

Chet

i get W=8.66N ..but how the answer is 5N

 

[haruspex]

i get W=8.66N ..but how the answer is 5N

Steady on, so far all we have found is the hanging weight. The tension in one string immediately after the other is cut might be different. So leave the weight in the form $10\sin(60)N$ for now.
Now draw a free body diagram for the weight just after one string is cut. What equation can you obtain for the tension in the remaining string.

 

[stuc]

Steady on, so far all we have found is the hanging weight. The tension in one string immediately after the other is cut might be different. So leave the weight in the form $10\sin(60)N$ for now.
Now draw a free body diagram for the weight just after one string is cut. What equation can you obtain for the tension in the remaining string.

the ans is 10N after cut the 5N right rope...huhu near but still cannot get the ans

 

[haruspex]

View attachment 91592

the ans is 10N after cut the 5N right rope...huhu near but still cannot get the ans

After a rope is cut, the system is no longer stable. There will be an acceleration. The forces in the vertical direction will no longer balance.
You need to consider which way that acceleration will be and apply $\Sigma F=0$ at right angles to it.

 

[stuc]

After a rope is cut, the system is no longer stable. There will be an acceleration. The forces in the vertical direction will no longer balance.
You need to consider which way that acceleration will be and apply $\Sigma F=0$ at right angles to it.

if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 10sin(60)N load on it.Is it my assumption is correct?

it mean all the calculation in step 25 not require especially step 2 &3 ?

 

[haruspex]

if one of the ropes have been cut, it mean the object just hang VERTICALLY with one rope with 10sin(60)N load on it.Is it my assumption is correct?

Not immediately. Although the question does not make it completely clear, I am very confident that you are supposed to find the tension in the remaining rope the instant after one rope is cut. You can take the remaining rope and the mass as being in exactly the same positions as before, and, for the instant, stationary. But now there is a nonzero acceleration.
Which way will the mass accelerate?

 

[stuc]

Not immediately. Although the question does not make it completely clear, I am very confident that you are supposed to find the tension in the remaining rope the instant after one rope is cut. You can take the remaining rope and the mass as being in exactly the same positions as before, and, for the instant, stationary. But now there is a nonzero acceleration.
Which way will the mass accelerate?

Do you mean the answer is wrong (Given ans: 5N)..i still stuck

 

[haruspex]

Do you mean the answer is wrong (Given ans: 5N)..i still stuck

Yes, it's wrong. (That would be the right answer if the angle were 45 degrees.)
You have established that the weight is 10sin(60) N. Draw a free body diagram, exactly like the original diagram, but with only one rope.
So the rope is still at 60 degrees to the horizontal.
Let the tension in the rope be T. You know the weight, and hence the mass. The tension and the weight are the only forces acting, agreed?
Which way will the mass accelerate?
What can you therefore say about the forces orthogonal to that direction?