# Magnitude of Force on Wrench

by dragon18
Tags: force, magnitude, wrench
 P: 12 You are installing a new spark plug in your car, and the manual specifies that it be tightened to a torque that has a magnitude of 45.0 N*m. Using the data in the figure below (L = 0.300 m and ? = 52.6°), determine the magnitude F of the force that you must exert on the wrench. Relevant Equations T = F * d Attempt 45 = F * cos 52.6º * 0.300m F = 246.96 N
 P: 31 What if θ was zero? According to your formula, this would give a torque of Fd (because cos(0) = 1). But, if you look at the picture, if θ = 0, shouldn't the torque be 0 as well? You are just pulling straight out; this shouldn't turn the bolt at all. Hint: is θ really the angle between the force and the displacement?
P: 12
 Quote by Poley What if θ was zero? According to your formula, this would give a torque of Fd (because cos(0) = 1). But, if you look at the picture, if θ = 0, shouldn't the torque be 0 as well? You are just pulling straight out; this shouldn't turn the bolt at all. Hint: is θ really the angle between the force and the displacement?
If θ=0, then the force would be 150N. θ isn't the angle between the force and displacement, but what am I looking for instead of 52.6°?

P: 31

## Magnitude of Force on Wrench

 Quote by dragon18 If θ=0, then the force would be 150N. θ isn't the angle between the force and displacement, but what am I looking for instead of 52.6°?
Basically, you want the component of F that is PERPENDICULAR to the displacement. Personally, I wouldn't use cosine but sine. Draw a right triangle with theta, F, and d, and find an expression for the component of F pointing away from d.
 P: 12 I figured out that the answer is 1.898102N F=45F*m/(0.300m*sin(52.6°))

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