# Magnitude of Force on Wrench

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

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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?

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°?

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.

I figured out that the answer is 1.898102N
F=45F*m/(0.300m*sin(52.6°))