Calculating Force for Spark Plug Installation

AI Thread Summary
To calculate the force required for spark plug installation, the torque specified is 45.0 N*m, with a lever arm length of 0.300 m and an angle of 52.6°. The initial calculation using the cosine of the angle yielded a force of 246.96 N, but this raised questions about the angle's relevance to torque. It was clarified that the angle should actually represent the perpendicular component of the force relative to the displacement, suggesting the use of sine instead of cosine. Ultimately, the correct force calculation led to a value of approximately 1.898 N when using the sine function for the angle. Understanding the relationship between force, angle, and torque is crucial for accurate spark plug installation.
dragon18
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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.
chaning your spark plugs.png



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?
 
Poley said:
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°?
 
dragon18 said:
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°))
 
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