What is the Sum of Torques on a Nut Placed in a Nutcracker with One Lever?

[warfreak131]

Homework Statement

The nut is now placed in a nutcracker with only one lever, as shown, (Intro 3 figure) and again friction keeps the nut from slipping. The top "jaw" (in black) is fixed to a stationary frame so that a person just has to apply a force to the bottom lever. Assume that $$F_2$$ is directed perpendicular to the handle.

Homework Equations

$$\Sigma\tau=0$$

The Attempt at a Solution

I'm not sure if the sum of the torques includes $$F_{2}D$$ and/or $$F_{2}d$$. I though that it was the sum of the torque of the smaller handle + the torque of the larger handle + the normal force. Since counter clockwise is positive, it would be $$\tau_{nut}-\tau_{small}-\tau_{large}=0$$ and solve for $$F_{2}$$, but that's not right.

 

[stewartcs]

Homework Statement

The nut is now placed in a nutcracker with only one lever, as shown, (Intro 3 figure) and again friction keeps the nut from slipping. The top "jaw" (in black) is fixed to a stationary frame so that a person just has to apply a force to the bottom lever. Assume that $$F_2$$ is directed perpendicular to the handle.

Homework Equations

$$\Sigma\tau=0$$

The Attempt at a Solution

I'm not sure if the sum of the torques includes $$F_{2}D$$ and/or $$F_{2}d$$. I though that it was the sum of the torque of the smaller handle + the torque of the larger handle + the normal force. Since counter clockwise is positive, it would be $$\tau_{nut}-\tau_{small}-\tau_{large}=0$$ and solve for $$F_{2}$$, but that's not right.

What is the problem asking you to find?

CS

 

[Tzim]

I think that the torques are two.The one from the force on the large part and the one from the nut.correct me please if i am wrong

 

[Tzim]

the jaw is fixed so the torque at it does not count

 

[warfreak131]

The question asks to find $$F_{2}$$. Since its the max torque just to begin to break the shell, the sum of all the torques would be 0. So wouldn't that mean that the normal force of the shell at distance $$d$$ equals the torque of the little lever plus the torque of the larger lever?

 

[Tzim]

I believe that the torque due to the shell on the small lever should be equal to F2*D

 

[Tzim]

if it is just to start break the F2*D>=Torque from the nut

 

[stewartcs]

The question asks to find $$F_{2}$$. Since its the max torque just to begin to break the shell, the sum of all the torques would be 0. So wouldn't that mean that the normal force of the shell at distance $$d$$ equals the torque of the little lever plus the torque of the larger lever?

Is this a continuation of another problem? If so, what else are you give (or did you find)?

The torque on the hinge would be equal to T = F_2 x D since it is the only force being applied. The perpendicular force acting on the nut would then be, F_nut = T / d since the torque at the hinge is the same.

CS

 

[warfreak131]

the correct answer according to our online physics homework is $$\frac{F_{n}d}{D}$$