Checking Basic Physics Problems: Torque, Balance, and Work

  • Thread starter BunsenBurner1
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In summary: I have been on both sides. Being explicit is important.In summary, the conversation discusses three problems involving torque, balance, and work. The formulas used for each problem are correct but there is some ambiguity in the wording of the first problem. The concept of energy is also mentioned and clarified as being related but not equivalent to work.
  • #1
BunsenBurner1
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[SOLVED] correct me if I'm wrong please

just a couple of problems i want to make sure i did correctly.

1) A force of 4.0 N is applied to a door at an angle of 60 degrees and a distance of .30 m from the hinge. What is the torque produced?
T=F*d*sin theta --> 1.04 Nm

2) A child with a weight of 4.5 x 10^2 N sits on a seesaw .60 m from the axis of rotation. How far from the axis of rotation on the other side should a child with a weight of
6.00 x 10^2 N sit so the seesaw will remain balanced?
m1gd1=m2gd2 --> .45 m

3) What work is required to lift a 500 kg mass a distance of 9 meters using a machine that is 65% efficient?
Efficiency=output/input --> .65= 4900(9)/input --> 6.78 x 10^4 J

i think these were simple but I'm just double checking. thanks guys.
 
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  • #2
The formulas are correct; I haven't checked the values.

The last problem is not properly framed. Does it mean how much energy the machine has to output in order to lift it?
 
  • #3
well isn't Energy and Work equivalent concepts?
E = W = Fd?
 
  • #4
BunsenBurner1 said:
well isn't Energy and Work equivalent concepts?
E = W = Fd?

I would say related, not equivalent. For the problem you're working on, W = Fd is fine. The concept of energy is a bit more complicated than its connecting to work alone.

As for your problems, (2) and (3) are fine. In (1), if we are to take it that the 60º angle is the angle the applied force makes to the plane of the door, then that answer is also fine; the wording of the problem leaves this unclear, though.
 
  • #5
dynamicsolo said:
In (1), if we are to take it that the 60º angle is the angle the applied force makes to the plane of the door, then that answer is also fine; the wording of the problem leaves this unclear, though.

The force can be applied in many ways and still make a angle of 60 deg with the plane of the door. To word it more precisely (assuming the door is in a vertical plane), we have to take the applied force to be in the horizontal plane, and 60 deg should be the angle between the force and the line this horizontal plane makes by intersecting the plane of the door.
 
  • #6
Shooting star said:
The force can be applied in many ways and still make a angle of 60 deg with the plane of the door. ...

Quite so, but my issue was with the phrase "a force ... is applied to a door at an angle of 60 degrees" : 60º relative to what? When torque problems are usually presented, there is somewhat more care in clarifying how the angles are to be measured. (My alternate answer was T = (4 N)(0.3 m)(sin 30º)...)
 
  • #7
dynamicsolo said:
(My alternate answer was T = (4 N)(0.3 m)(sin 30º)...)

You mean you thought that 60 deg was the angle of the force with the normal to the plane?
 
  • #8
Shooting star said:
You mean you thought that 60 deg was the angle of the force with the normal to the plane?

I'm saying the statement of the problem doesn't say how the 60º is measured. The usual intention would be that it is to be measured from the plane of the door, but the wording (as posted) doesn't actually say this, so a student could be forgiven for coming up with the other answer. (Having had to deal with problem-writing and problem-grading on and off for a long time, I've found one has to be very sensitive to how a problem is worded. I've run into plenty of situations where alternative answers had to be accepted because the person writing the problem -- occasionally, me! -- had not been explicit enough.)
 
  • #9
I agree with you whole heartedly.
 

Related to Checking Basic Physics Problems: Torque, Balance, and Work

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