Determine the magnitude of force on a suspended object

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Homework Help Overview

The problem involves a 5.0 kg mass suspended by a rope with a horizontally directed force applied to it, aiming to determine the magnitude of that force to achieve a specific angle of 65 degrees. The discussion centers around the application of physics principles related to forces and angles in a static equilibrium context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss various approaches to solving the problem, including the use of free-body diagrams and trigonometric relationships. There are questions about the application of specific formulas and the relevance of torque in this context. Some participants express uncertainty about the correct interpretation of the angle and its implications for the calculations.

Discussion Status

The discussion is ongoing with participants providing feedback on each other's reasoning and calculations. Some guidance has been offered regarding the use of diagrams and the importance of understanding the forces acting on the object. There is a recognition of the need to clarify the angle's reference point, and participants are exploring different interpretations and calculations without reaching a definitive conclusion.

Contextual Notes

There is a mention of missing information, such as the length of the rope, which affects the ability to apply certain formulas. Additionally, participants are navigating the ambiguity of the angle's reference point, which could influence the calculations and understanding of the problem.

MadMcB
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Homework Statement


A 5.0 kg mass is suspended by a rope. A horizontally directed force F is applied to the mass. What magnitude of force is needed to produce an angle of 65˚?

mass = 5kg
angle = 65˚

Homework Equations



F = mgsinx
T = rFsinx

The Attempt at a Solution


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I tried using the first formula:
F = (5kg)(.8m/s^2)(sin65˚)
F = 44N

I'm not sure if this is correct though, I am worried I used the wrong formula, but since I don't know the length of the rope I can't use the formula for torque.
 
Last edited:
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Hello MadMcB. Welcome to PF!

Your answer is not correct. Be careful with pulling formulas out of your notes unless you are sure they apply to your new problem. You can solve this problem without using any physics formulas, just a little trig.

When the object is being held at rest at 65o, how many forces act on the object? Draw a force diagram ("free-body diagram") and proceed from there.
 
You shouldn't use torque. You know the magnitude of the net force and it's vertical component.

(Although you calculated F incorrectly. Try drawing a diagram rather than using a formula, formulas don't always work because details may change)

With this, what can you solve for?
 
TSny said:
Hello MadMcB. Welcome to PF!

Your answer is not correct. Be careful with pulling formulas out of your notes unless you are sure they apply to your new problem. You can solve this problem without using any physics formulas, just a little trig.

When the object is being held at rest at 65o, how many forces act on the object? Draw a force diagram ("free-body diagram") and proceed from there.
Hi TSny! Thanks for the advice. I've drawn a diagram, and I know that the vertical component of the normal force on the object is 49N. How do I find the vertical component?
 
Brian T said:
You shouldn't use torque. You know the magnitude of the net force and it's vertical component.

(Although you calculated F incorrectly. Try drawing a diagram rather than using a formula, formulas don't always work because details may change)

With this, what can you solve for?
Thanks Brain. How do I know the net force? If I had the net force and the vertical component, I could solve for the horizontal component...
 
Could you perhaps attach an img of your diagram so we could see what you have?
 
Brian T said:
Could you perhaps attach an img of your diagram so we could see what you have?
Can I insert an image as a file from my computer??
 
Brian T said:
Could you perhaps attach an img of your diagram so we could see what you have?
here you go! Apologies- It is sideways
 

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Okay. Just one question: is it 65 degrees from the horizontal or from the perpendicular?

Once you figure that out (not sure if the problem makes it ambiguous), use trig to figure out the tension and then get the horizontal component from that
 
  • #10
Brian T said:
Okay. Just one question: is it 65 degrees from the horizontal or from the perpendicular?

Once you figure that out (not sure if the problem makes it ambiguous), use trig to figure out the tension and then get the horizontal component from that
65˚ from the horizontal. So the tension is sin65˚ = o/h = 49/h. So h = 49/sin65 = 54.07 N

using pythagorean's theorem: x^2 + 49^2 = 54.07^2. So x^2 = 522.56, and x = 22.86 N

So, F = 23N. Is this correct?
 
  • #11
MadMcB said:
65˚ from the horizontal. So the tension is sin65˚ = o/h = 49/h. So h = 49/sin65 = 54.07 N

using pythagorean's theorem: x^2 + 49^2 = 54.07^2. So x^2 = 522.56, and x = 22.86 N

So, F = 23N. Is this correct?

Yup looks good.
 
  • #12
Brian T said:
Yup looks good.
Thanks for your help.
 

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