Calculate Min Angle to Exert 35.0 NM Torque with 355 N Force

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To calculate the minimum angle required to exert a torque of 35.0 N·m with a force of 355 N on a 0.150 m wrench, the equation t = rfsinθ is used. Rearranging gives sinθ = τ/(rF), leading to θ = arcsin(τ/(rF)). Substituting the values results in a very small angle, indicating that a precise application of force is necessary. The discussion highlights confusion about rearranging the equation and applying the arcsin function correctly. Understanding the relationship between torque, force, and angle is crucial for solving this problem effectively.
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Homework Statement


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?


Homework Equations


t = rfsinθ


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
 
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lacar213 said:

Homework Statement


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?


Homework Equations


t = rfsinθ


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

\vec{\tau}=\vec{r}\times\vec{F}\rightarrow\frac{\vec{\tau}}{rF}=\sin(\theta)...
 
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??
 
lacar213 said:
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??

You apply the arcsin function to both sides of the equation, yielding:

\arcsin\left(\frac{\vec{\tau}}{rF}\right)=\theta.
 
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