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Am I doing this torque problem right?

  1. Oct 30, 2013 #1

    I drew up the picture in cad so that you can look at it. The find torque would i take the cross product of the length of the beam (10m) and the component of the force that is applied perpendicular to the beam (50Nsin45°)? I then get ≈353.55Nm and then use the right hand rule to find that the direction is out of the page (or screen).

    Is that all correct, cause another student said that I am supposed to multiply the horizontal component of the beam by the portion of the force perpendicular to the beam (10mcos45° * 50Nsin45) to find the magnitude of the torque. This seems backward to me cause that seems like a dot product and I thought that torque was defined as the cross product of the position vector and the force vector.
  2. jcsd
  3. Oct 30, 2013 #2


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    When calculating T = r x F, both r and F must be expressed in vector form. For your example, what are the r and F vectors? Forget about street physics. If you apply the correct formula with the correct data, you will get the correct value of the torque.
  4. Oct 30, 2013 #3
    I'm sorry, you lost me, it has been a couple of years since I have done none street physics honestly. But it is the cross product of r and F vectors then? Because this other student is saying that you need to use the perpendicular distance (magnitude) instead of r (both a vector sum and not restrained to just the vertical or horizontal axis) and simply multiply it by the force and that still seems wrong to me. So I think that you are saying my approach is correct yet and oversimplification of the mechanics at work???
  5. Oct 30, 2013 #4


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    You are right, he is wrong.
  6. Oct 30, 2013 #5
    alright, thank you. Now the same conversation but a new example, and he still insists that he is correct.
    A ladder that is 10ft long, weighs 30lbs, and has a 180 lb man standing 5ft up it, resting at 60 degrees on a wall. I would approach this problem by first calculating the torque from the stop at the floor

    180lbs cos60° * 2.5ft + 30lbs cos60° * 10ft = Fwall sin60° * 10ft (all trig functions used to find component of force that is perpendicular to the ladder)
    and when I solve for Fwall i get ≈43.3lbs

    then I would simply set the force of friction (or as he has in his problem, the person standing at the bottom of the ladder on a frictionless floor, god know how that works, lol) to that as they are the only horizontal forces, then the normal force from the ground would be the two weights added together. So Ffriction is also 43.3lbs and Fnormal is 210lbs.

    He claims that you would do the same except use this equation instead at the beginning
    180lbs cos60° * 2.5ft cos60° + 30lbs cos60° * 10ft cos60° = Fwall sin60° * 10ft sin60°
    Solve Fwall for 25lbs

    This does not make sense to me, yet he has actually showed this to me in his textbook. I have noticed that he always calls them moments instead of torques, but I thought that was just an old engineering convention but still meant the same thing as a torque. I have never heard of a dot product of a position vector and a force vector.


    I find this confusing as I was a physics major until I ran out of money and I never heard of this stuff, yet he has a book written by a Dr. Thomas Burns that says I am wrong. Where is the discrepancy?
  7. Oct 30, 2013 #6


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    Moments and torques are the same thing. 'Torque' is the preferred term when talking about the energy generated by an engine, for example.

    You can calculate moments as you have described in the example above, by getting the components of the forces acting normal to the ladder, or you can calculate the distances between forces measured along the floor. Taking moments about the foot of the ladder using these distances and the forces as shown in your second picture can be used to solve for the unknown reactions. It's just a matter of writing the proper equilibrium equations.

    Your problem can also be solved using the old standby T = r x F. The important thing is to learn a method thoroughly which can be used to solve a problem. There may be several superficially different ways to approach a solution, but it is not required that you learn all the variations in these methods.

    BTW, your equation doesn't match the diagram in the first picture: the weight of the ladder is centered at 5 feet from the end, not 10 feet. This mistake is also in your friend's equation.

    For Fwall, I get 34.6 lbs

    As long as ƩF = 0 and ƩM = 0, you are golden regardless of solution method.
  8. Oct 30, 2013 #7
    ok, thank you for clearing that up. And woops, didnt realize i accidentally wrote down 10 instead of 5, good job picking that up
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