Confused about resolving Tension and Weight

  • #1
laser
104
17
Homework Statement
Confused about resolving Tension and Weight
Relevant Equations
uhh
1700939433965.png


Calculations with 1:
T1sintheta + T2sintheta = W
T1costheta = T2costheta

Calculations with 2:
Wsintheta = T1
Wcostheta = T2

These are not equivalent. Can someone point out the flaw in my logic?

Edit: System is in equilibrium!
 
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  • #2
laser said:
Homework Statement: Confused about resolving Tension and Weight
Welcome to PF!
When giving the Homework Statement, please give the full statement exactly as given to you.

laser said:
Relevant Equations: uhh
Can you list any relevant equations for the forces when you have static equilibrium?

laser said:
Calculations with 1:
T1sintheta + T2sintheta = W
T1costheta = T2costheta
These look right.

laser said:
Calculations with 2:
Wsintheta = T1
Wcostheta = T2

These are not equivalent. Can someone point out the flaw in my logic?

It's hard to follow your logic based on the little that you have written down. The equation ##W \sin \theta = T_1## is incorrect. I'm guessing that you neglected the fact that ##T_2## has a component parallel to ##T_1##. That is, ##T_2## is not perpendicular to ##T_1## for general values of ##\theta##. Likewise, your equation ##W \cos \theta = T_2## is incorrect.
 
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  • #3
TSny said:
When giving the Homework Statement, please give the full statement exactly as given to you.
Oops, probably posted in the wrong forum. This isn't a homework, question, just something I was wondering about.
TSny said:
It's hard to follow your logic based on the little that you have written down. The equation Wsin⁡θ=T1 is incorrect. I'm guessing that you neglected the fact that T2 has a component parallel to T1. That is, T2 is not perpendicular to T1 for general values of θ. Likewise, your equation Wcos⁡θ=T2 is incorrect.
Fair point, I agree with you.

Let's say theta = 45 degrees. That makes them perpendicular, but the equations still don't work out.

From calculation 1:
We get W = Tsqrt(2)

From calculation 2:
We get W=T/sqrt(2)
 
  • #4
laser said:
Let's say theta = 45 degrees. That makes them perpendicular, but the equations still don't work out.

From calculation 1:
We get W = Tsqrt(2)

From calculation 2:
We get W=T/sqrt(2)
Check your equation for calculation 2.
 
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  • #5
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Related to Confused about resolving Tension and Weight

What is the difference between tension and weight?

Weight is the force exerted by gravity on an object, calculated as the product of the object's mass and the acceleration due to gravity (W = mg). Tension, on the other hand, is the force transmitted through a string, rope, cable, or any other form of a flexible connector when it is pulled tight by forces acting from opposite ends.

How do you calculate the tension in a rope holding a suspended object?

To calculate the tension in a rope holding a suspended object at rest, you need to consider the forces acting on the object. If the object is in equilibrium (not accelerating), the tension in the rope is equal to the weight of the object. Therefore, T = mg, where T is the tension, m is the mass of the object, and g is the acceleration due to gravity.

What happens to the tension in a rope if the object is accelerating?

If the object is accelerating, the tension in the rope will change depending on the direction of the acceleration. If the object is accelerating upwards, the tension will be greater than the weight of the object (T = mg + ma). If the object is accelerating downwards, the tension will be less than the weight of the object (T = mg - ma), where a is the acceleration.

How do you resolve tension in a system with multiple ropes or angles?

In a system with multiple ropes or angles, you need to resolve the tension into its components. This involves breaking down the tension forces into their horizontal and vertical components using trigonometric functions (sine and cosine). You can then apply equilibrium conditions (sum of forces in each direction equals zero) to solve for the unknown tensions.

Can tension in a rope be different at different points along its length?

In an ideal, massless rope, the tension is the same throughout its length. However, if the rope has mass, the tension will vary along its length due to the weight of the rope itself. The tension will be greatest at the point where the rope is attached to the support and will decrease as you move towards the end of the rope.

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