Mastering Lever Mechanics for MCATs to Understanding Net Torques

[cxz]

Hi guys. I haven't had physics in a couple of years and I'm trying to study for the MCATs. I've always had trouble with levers so I would really appreciate it if I could get some help. In particular, the following scenario confuses me:

Say you have a man standing on a board, the fulcrum is to the left of the man's center of gravity, and he is pulling up on the right edge of the board. It is to my understanding that in such scenarios, the upward force applied by the man's arm is irrelevant towards calculations of net torque around the fulcrum because of the downward reactionary force applied by his feet, and that the net torque on the fulcrum is determined purely by the man's weight at his center of gravity.

This makes sense to me intuitively, but when I try to work out the numbers, I confuse myself. Let's say the board's length is 10m, the fulcrum is located all the way at the left end, the man's center of mass is 1m from the left, and the man is pulling up on the rightmost edge of the board. I'm assuming that the reactionary force exerted by his feet on the board is equal and opposite to the upward force exerted by his arms, which we will call F. Taking counter-clockwise rotation to be positive, the net torque around the fulcrum can then be calculated as:
torque = F(10) - F(1) - mg(1) = F(9) - mg(1)

This equation is obviously wrong because it suggests that in such a scenario, the man is able to lift himself up off the floor. Essentially, to take filght. I know I made a mistake somewhere. Please help.

 

[mgb_phys]

The reaction force is his weight PLUS the reaction force from his arms.

 

[cxz]

Hmmm do you mean that the equation should look like this?

torque = F*10 - [ ( F*1 + mg*1) + mg*1 ] = F*9 - mg*2?

If so, then if the distance between his arms and his center of mass is long enough, wouldn't he still be able to lift himself up?

 

[mgb_phys]

Sorry wasn't thinking straight -
The force he pulls up with his arms must equal the force he pushes down with his feet (which is partly weight and partly reaction).
The up and down forces he exterts must balance or he would fly off.
But there is a reaction force from the plank at the point his feet make contact and the point his arms make contact.
So considering the torques from the fulcrum:
T=(force from feet * distance1) + (reaction to feet * distance1) + (force arms * distance2) + (reaction arms * distance2).
Since (force feet) = - (reaction feet) and (force arms)=-(reaction arms) there is no net torque and he can't pull himself up by his own bootstraps!

 

[cxz]

Hmm sorry to be a bother but I'm still kind of confused about the equation you gave. Assuming that the torque you are referring to is the torque on the board, then I don't see how there can be a reaction force from the arms and legs of the person acting on the board. The only forces that the board exerts on the person are reactionary forces. How can the person exert more reactionary forces on the board in reaction to the reactionary forces exerted on him by the board? That seems like a loop to me. It seems to me that you are using the free body diagram of the person in addition to the free body diagram of the board to calculate the torque on the board. Sorry, still confused.

 

[mgb_phys]

First think of the person just standing on a board. There is a reaction force at his feet balancing his weight - otherwise he would be moving.
Becuse the force down (weight) and the force (up) balance and act at the same point there is no net torqe.

 

[nealh149]

mqb_phys, one of those forces act on the board and the other act on the man, so the net torque due to his weight is definitely not zero