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Torque on Wrench

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data
    You want to exert a torque of at least 35.0 N·m on a wrench whose handle is 0.150 m long. If you can provide a force of 355 N to the end of the wrench, what is the minimum angle at which you can apply the force in order to achieve the desired torque?


    2. Relevant equations
    t = rfsinθ


    3. The attempt at a solution
    I am not sure how to rearrange this equation to find the angle or what other equation I have to use to find the angle.
    torque = 35.0 NM
    F = 355 N
    M = .150 M
     
  2. jcsd
  3. Nov 12, 2008 #2
    [tex]\vec{\tau}=\vec{r}\times\vec{F}\rightarrow\frac{\vec{\tau}}{rF}=\sin(\theta)[/tex]...
     
  4. Nov 12, 2008 #3
    After you divide the torque by the length and force do you take the sin of that answer - because when you do it is very small??
     
  5. Nov 12, 2008 #4
    You apply the arcsin function to both sides of the equation, yielding:

    [tex]\arcsin\left(\frac{\vec{\tau}}{rF}\right)=\theta[/tex].
     
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