• Support PF! Buy your school textbooks, materials and every day products Here!

Where did I go wrong?

  • #1
[Mentor's Note: Thread moved from the technical forums after receiving replies]
proxy.php?image=http%3A%2F%2Fi58.tinypic.com%2F24vsbgg.gif

I'm just learning torque, but if I went wrong here can someone help me (or explain the textbook explanation better).

My work was:
Torque=.4m(weight of 5kg)sin(120)
which comes out to 1.73
The answer choices are
A=.49
B=.98
C=1.7
D=2
E=3.4

The answer is D, and it is apparently because "from the diagram, sin(theta)=40cm/L, so Lsin(theta)=40cm. That is, the question gave the lever arm already. (The given angle of 60 degrees is irrelevant.) Thus, from Torque=Fl, the torque is (.4m)(.5kg)(10m/s^2)=2."

Why doesn't Torque=rFsin(theta) apply? Why is the angle irrelevant?

Thank you to all who reply; this is my first post so if I make any mistakes I apologize.
 
Last edited by a moderator:

Answers and Replies

  • #2
3,732
414
What is the question?
 
  • #3
Oh my gosh I'm so sorry.

"what is the torque about the pendulum's suspension point produced by the weight of the bob?"
 
  • #4
3,732
414
Just this? Nothing about the mass and the length of the pendulum?
 
  • #5
I should have included that in the question also.
mass=.5kg and is 40cm below the suspension point. No length is given. I suppose that's where I messed up?
 
  • #6
3,732
414
It is really painful to write the exact text? :smile:
What is 40 cm below suspension point? The mass?

It may seem funny, but the answer to a question depend on the details of that question.
 
  • #7
My apologies.

"In the figure above, what is the torque about the pendulum's suspension point produced by the weight of the bob, given that the mass is 40cm below the suspension point, measured vertically, and m=.50kg?"
 
  • #8
3,732
414
Ok, now it makes sense. :smile:
No, the lever arm is not already given. And you will need the angle to find it.
40 cm is the vertical distance. Imagine that you draw a vertical line from m upwards, until it reaches the thick horizontal line.
You will have a right angle triangle, with the vertical side 40 cm.
The lever arm of the weight is the horizontal side of that triangle.

Unless they mean that 40 cm refers to the equilibrium position of the mass, in which case 40 cm will be the length of the pendulum.
 
Last edited:
  • #9
Two questions:

What is done from there?
If the length of the string was 40 cm would my original attempt be correct?
 
  • #10
3,732
414
The formula for torque applies, of course.
The angle in the formula is between r (vector from origin) and the force.
Can you see what this angle is in your case?

I think their answer may not be correct.

Edit
I did not see that you wrote that first formula, with sin 120.
If the length of the pendulum were 40 cm, it will make partial sense. The angle however is not 120. I see why you may think it is so. But to see the actual angle, imagine that you put both vectors with the tail in the origin.
 
Last edited:
  • #11
Isn't the angle 120 in my case?
 
  • #12
3,732
414
I edited my last post while you were posting. Sorry. See above.

Edit (again)
If the length of the pendulum is 40 cm, then the answer will be on the list but not D.
You just need to take correct angle in your first formula and you will get it.
 
Last edited:
  • #13
So the angle would be thirty, and their answer is wrong? (An angle of 30 yields an answer of exactly one, and since you said it is one of the answers I assume my angle is still off?) By the way, thank you so much for your patience and time
 
  • #14
3,732
414
Yes, if you take g=10 m/s^2.
With g=9.8 m/s^2 you get 0.98 Nm which is answer B.

And yes, the angle is 30 degrees.
 
  • #15
Oh, very well. The book has been using 10 so there is the mix-up. Thank you again
 
  • #16
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,532
4,974
Seems to me the author got confused between sine and cosine.
The text states correctly that ##L\sin(\theta)=.4m##, theta being the 60 degrees shown, but mistakenly takes ##L\sin(\theta)## as the lever arm instead of ##L\cos(\theta)##
 

Related Threads for: Where did I go wrong?

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
1
Views
1K
Replies
5
Views
10K
Replies
1
Views
877
Replies
7
Views
14K
Top