Physics Fundamentals : Free Body Diagram

  • #1

Homework Statement


A 5.0kg mass is suspended from the ceiling. A horizontal force F is applied to hold the string in position as shown

http://i.imgur.com/ZdR72ff.jpg

Determine the reading of F on the spring balance by using a scale diagram

Homework Equations



None

The Attempt at a Solution



http://i.imgur.com/2KO2AnT.jpg
Am I doing it correctly?
It doesn't really make sense to me. I can't determine the force F. I only have 30degree angle and the weight.
I don't know how to draw the length of Force F. What is it should I be looking for?

Thank you!
 

Answers and Replies

  • #2
phinds
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Have you studied force vectors?
 
  • #3
Andrew Mason
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Homework Statement


A 5.0kg mass is suspended from the ceiling. A horizontal force F is applied to hold the string in position as shown

http://i.imgur.com/ZdR72ff.jpg

Determine the reading of F on the spring balance by using a scale diagram

Homework Equations



None

The Attempt at a Solution



http://i.imgur.com/2KO2AnT.jpg
Am I doing it correctly?
No.

It doesn't really make sense to me. I can't determine the force F. I only have 30degree angle and the weight.
I don't know how to draw the length of Force F. What is it should I be looking for?
1. What are all the forces are on the 5.0 kg. mass? Hint: I can see two tensions and gravity. Are there any others?

2. Draw force vectors for each force. hint: You have to work out the magnitude of the tensions, so put a "?" for them for now. You can determine the force and direction of gravity.

3. What do the forces all sum to? (hint: is there any acceleration here?).

4. Work out the magnitidue of the unknown forces.

AM
 
  • #4
Have you studied force vectors?
I have look through my textbook and for this question I am having weird answers. I believd I have not fully understand yet.


No.


1. What are all the forces are on the 5.0 kg. mass? Hint: I can see two tensions and gravity. Are there any others?

2. Draw force vectors for each force. hint: You have to work out the magnitude of the tensions, so put a "?" for them for now. You can determine the force and direction of gravity.

3. What do the forces all sum to? (hint: is there any acceleration here?).

4. Work out the magnitidue of the unknown forces.

AM
I have tried to identify all the forces.

http://i.imgur.com/HexCRVQ.jpg

Does this make sense? I made a bad calculation mistake on previous attempt
From what I understand, mass will exert a weight of 50N. Therefore the tension at T1 will be 50N. Tension at T2 will also be 50N because the weight is pulling it down. The resultant force has to be 0 as it is stationary. The force on the weight and tension will cancel out each other. I am now left with Force F. I can't go anywhere from here. Which part am I getting it wrongly?

Thank you!
 
  • #5
Doc Al
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From what I understand, mass will exert a weight of 50N. Therefore the tension at T1 will be 50N.
Good.

Tension at T2 will also be 50N because the weight is pulling it down.
No. Realize that T2 acts at an angle.

The resultant force has to be 0 as it is stationary. The force on the weight and tension will cancel out each other. I am now left with Force F. I can't go anywhere from here. Which part am I getting it wrongly?
T2 does not cancel the weight. But the vertical component of T2 does.
 
  • #6
Good.


No. Realize that T2 acts at an angle.


T2 does not cancel the weight. But the vertical component of T2 does.
OH!
If I am understanding it correctly,
The vertical component of T2 is 50N
Therefore force F is at the base.

http://i.imgur.com/NCCD8Hy.jpg

Am I right to say:
Using a scale diagram, 1cm representing 10N


sin(30) = F/5
F = 5sin(30)
F= 2.5cm

Therefore, the reading of force F is 25N.

Am I understanding it correctly?

Thank you!
 
  • #7
Nathanael
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sin(30) = F/5
F = 5sin(30)
F= 2.5cm
Why did you say [itex]sin(30°) = \frac{F}{5}[/itex] ?


Wouldn't it be [itex]sin(30°) = \frac{F}{T_2}[/itex]



("sine = opposite / hypotenuse")
 
  • #8
Why did you say [itex]sin(30°) = \frac{F}{5}[/itex] ?


Wouldn't it be [itex]sin(30°) = \frac{F}{T_2}[/itex]



("sine = opposite / hypotenuse")
Oh my! Sorry! What a silly mistake I made.

It should be
tan(30) = F/5 (opposite/adjacent)
F = 5tan(30)
F = 2.89 (3 sig. fig.)

Therefore the force F is 28.9N

Am I understanding this correctly?

Thank you!
 
  • #9
Nathanael
Homework Helper
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Oh my! Sorry! What a silly mistake I made.

It should be
tan(30) = F/5 (opposite/adjacent)
F = 5tan(30)
F = 2.89 (3 sig. fig.)

Therefore the force F is 28.9N

Am I understanding this correctly?

Thank you!
Yes I think that is correct.
 
  • #10
Yes I think that is correct.
Thank you very much for your help!

So my mistake was not recognising the vertical component of T2.

Thank you!
 

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