Torque Of Lever Arm in 1st/2nd Class Levers

[EnquiringMind]

Part 1 - Question About Torque Of Lever Arm in a 1st Class Lever

Homework Statement

A little while back I posted a question about incorporating the torque of a uniform mass and dimension lever arm into the principle of a 1st class lever.

Here is the thread for reference.
https://www.physicsforums.com/showthread.php?t=340234"

I gave an answer that was deemed correct, but I am curious if I am supposed to incorporate the angle the lever is at as well?

Does the angle the lever is at (e.g. left side down low under a rock, right side high up in the air waiting for force to be applied) effect the calculation of torque in a first class lever?

Homework Equations

Does the angle the lever is at (e.g. left side down low under a rock, right side high up in the air waiting for force to be applied) effect the calculation of torque in a first class lever?

The Attempt at a Solution

I'm not familiar enough with physics to understand how to incorporate angles/"sin" into these equations.

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Part 2 - Question About Torque Of Lever Arm in a 2nd Class Lever

Homework Statement

How is the torque of a lever arm (uniform mass, uniform dimension) calculated and incorporated into the "equation/math" of a 2nd class lever?

I have a 12 foot, 200 lb. second class lever arm of uniform mass and dimension.

My fulcrum is attached to the ground, and underneath the lever at 2 feet from the fulcrum is a car that I am trying to smash. Basically, I have a nutcracker type second class lever that uses a lever arm and the ground.

The Attempt at a Solution

Is the Torque of this Lever Arm (6 foot x 200 lb.) 1200 foot lbs. which is added to the effort force when trying to generate enough "down force" at 2 feet to adequately crush the car?

Does the starting angle of the lever matter in calculating torque in this case?-----------------I appreciate any feedback.

 

[rl.bhat]

Q1) In the lever problem, if the fulcrum is not in the midpoint of the bar, find the distance between the midpoint and fulcrum and find the monent due to weight of the bar about the fulcrum.
Torque is always F*r*sinθ, θ is the angle between force and bar.

 

[tomcue]

Thank you for your questions about torque in first and second class levers. In both cases, the torque of the lever arm is calculated by multiplying the force applied to the lever by the distance from the fulcrum to the point where the force is applied. This calculation does not take into account the angle of the lever arm.

For a first class lever, the angle of the lever arm does not affect the calculation of torque. This is because the fulcrum is located between the force and the load, so the direction of the force is always perpendicular to the direction of the lever arm.

For a second class lever, the angle of the lever arm does play a role in the calculation of torque. In this case, the load is located between the force and the fulcrum, so the direction of the force is not perpendicular to the lever arm. The torque is still calculated by multiplying the force by the distance from the fulcrum to the point where the force is applied, but the angle of the lever arm must also be taken into account. This can be done using the sine function, which relates the angle of the lever arm to the distance from the fulcrum.

In your example, the torque of the lever arm would be 1200 foot-pounds, and this would be added to the force applied at the 2-foot mark. The starting angle of the lever arm does matter in this case, as it will affect the amount of force needed to generate enough "down force" to crush the car.

I hope this helps clarify the role of torque in first and second class levers. If you have any further questions, please don't hesitate to ask.