Show that the sum of the torques at the given point is zero

[VitaX]

Homework Statement

Homework Equations

Torque = r*F

The Attempt at a Solution

I've tried multiple times now to get the sum of the torques at the given points equal to zero but I can't figure out a way to do it. It just doesn't come close to becoming zero.

Edit: Forgot to add that the fulcrums location is at .6 Meters

 

[berkeman]

Homework Statement

Homework Equations

Torque = r*F

The Attempt at a Solution

I've tried multiple times now to get the sum of the torques at the given points equal to zero but I can't figure out a way to do it. It just doesn't come close to becoming zero.

Edit: Forgot to add that the fulcrums location is at .6 Meters

Can you show us your calculations of the torques at those 3 points? What is the net torque about the fulcrum?

 

[VitaX]

Torque at 0 Meters = .1925*1.47+2.45*.3-2.666*.6 + 5.39*.9

Torque at .5 Meters = -.3075*1.47+.2*2.45+5.39*.4

Torque at 1 Meter = 5.39*.1-2.666*.4+2.45*.7+.8075*1.47

 

[berkeman]

Torque at 0 Meters = .1925*1.47+2.45*.3-2.666*.6 + 5.39*.9

Torque at .5 Meters = -.3075*1.47+.2*2.45+5.39*.4

Torque at 1 Meter = 5.39*.1-2.666*.4+2.45*.7+.8075*1.47

The weight pushing down on the fulcrum (and being opposed by the force up by the fulcrum) is not just the weight of the meter stick...

 

[VitaX]

The weight pushing down on the fulcrum (and being opposed by the force up by the fulcrum) is not just the weight of the meter stick...

Well how would you write it because I've tried literally every possibility I could think of and all the students I talked to in my physics lab have the same problem even after talking to the physics lab instructor. We're just more confused.

 

[berkeman]

Well how would you write it because I've tried literally every possibility I could think of and all the students I talked to in my physics lab have the same problem even after talking to the physics lab instructor. We're just more confused.

What is the total weight pushing down on the fulcrum?

 

[VitaX]

Total weight is the weight of all the hangers on the meter stick and the meter stick itself. But I have to have some negative values in my calculations somewhere else I won't get a value anywhere close to zero. The problem is what is negative.

 

[berkeman]

Total weight is the weight of all the hangers on the meter stick and the meter stick itself. But I have to have some negative values in my calculations somewhere else I won't get a value anywhere close to zero. The problem is what is negative.

Well, in the sum about 0, try adding in the total weights plus the stick for your one negative torque term... the total gets closer to zero, but is not zero yet. I think next you need to account for some of the stick weight in the torque at each weight position...

 

[VitaX]

I don't understand what you mean when you say "stick weight at each weight position". How would you even know what that would be. You can only say the stick weight at the center of mass or fulcrum I thought.

Here's a question when implementing the mass of the meter stick. Is this right: 2.666*.6 or is it going to be 2.666*.5 where .5 is center of mass or is it going to be 2.666*.1 where .1 is distance from center of mass to fulcrum. These little things are confusing me here. I need some clarification before I can write down another equation and post on here since mine are all off.

 

[vela]

It's going to be 2.666 N multiplied by the distance from the center of the meter stick to the point about which you are calculating the torques. If you're calculating about the left end of the meter stick, this distance would be 0.5 m.

What did you find for the force at the fulcrum?

 

[VitaX]

Friend told me to use these:

Torque at 0 cm = 2.666*.5 + 1.47*.1925 + 2.45*.3 + 5.39*.9 - .6(2.666 + 1.47 + 2.45 + 5.39) = 0.016 N
Torque at 50 cm = 2.666*.5 + 1.47(1-.1925) + 2.45*.7 + 5.39*.1 - .4(2.666 + 1.47 + 2.45 + 5.39) = - 0.016 N*m
Torque at 100 cm = 1.47(.5 - .1925) + 2.45*.2 - 5.39*.4 + .1(2.666 + 1.47 + 2.45 + 5.39) = - 0.016 N*m

Guess I'm fine now.