What is the torque needed to lift a weight with a lever?

[Priyadarshini]

Homework Statement

Homework Equations

Torque=one of the forces x the perpendicular distance
moment=force x perpendicular distance

The Attempt at a Solution

moment of pulling the weight up around the disc:
1.2/2 = 0.6 m
0.6-0.2 = 0.4 m
moment= 0.4 x 900 = 360 Nm
To lift it, the torque of the rod has to be 360 Nm too
So,
F x 1.2 = 360
F = 300 N

But the answer is 150N

 

[Buzz Bloom]

But the answer is 150N

Hi Priyadarshini:

Your description of your solution omits some details, so I am not sure where you went astray. I suggest thinking about the problem in terms of mechanical advantage.
What is the mechanical advantage of the 2F force (1F on each end of the lever) applied at its radius compared with the 900N force applies at the radius if the disc?

Hope this helps.

Regards,
Buzz

 

[Priyadarshini]

Hi Priyadarshini:

Your description of your solution omits some details, so I am not sure where you went astray. I suggest thinking about the problem in terms of mechanical advantage.
What is the mechanical advantage of the 2F force (1F on each end of the lever) applied at its radius compared with the 900N force applies at the radius if the disc?

Hope this helps.

Regards,
Buzz

Hi,
What exactly do you mean by mechanical advantage?

 

[CWatters]

Try calculating the torque on the vertical shaft due to the 900N force. The two forces F must create a torque at least as great.

 

[physics user1]

Hi, finding the minimum force is the same as saying that the torque of the disk has the same magnitude of the torque of the lever

T (lever) = F*d where d= 1.20 m
T (disk)= W*R where R is the radius of the disk and W is the weight

T (lever)= T (disk)

 

[haruspex]

0.6-0.2 = 0.4 m

This does not calculate anything useful. What physical law are you planning to use this in?
It's the ratio of the two distances that's interesting, not the difference.

 

[Buzz Bloom]

What exactly do you mean by mechanical advantage?

Hi Priyadarshini:

Please take a look at
https://en.wikipedia.org/wiki/Mechanical_advantage

Regards,
Buzz

 

[Priyadarshini]

Try calculating the torque on the vertical shaft due to the 900N force. The two forces F must create a torque at least as great.

Isn't that 0.4*900?

 

[Priyadarshini]

Hi, finding the minimum force is the same as saying that the torque of the disk has the same magnitude of the torque of the lever

T (lever) = F*d where d= 1.20 m
T (disk)= W*R where R is the radius of the disk and W is the weight

T (lever)= T (disk)

why do we take the radius of the disc instead of half the rod minus the radius of the disc? Isn't the turning point on the circumference of the disc where the string starts to rotate around the disc?

 

[Priyadarshini]

This does not calculate anything useful. What physical law are you planning to use this in?
It's the ratio of the two distances that's interesting, not the difference.

I thought that 0.4 was the perpendicular distance from the pivot for calculating the torque of the disc, which is why I subtracted.

 

[haruspex]

why do we take the radius of the disc instead of half the rod minus the radius of the disc? Isn't the turning point on the circumference of the disc where the string starts to rotate around the disc?

No, both rod and disc rotate about an axis through their centres.
I suppose you could consider that the not-yet-wound string instantaneously rotates about the point of contact with the disc, but since it exerts no torque about that point it is quite irrelevant.

 

[Priyadarshini]

Hi Priyadarshini:

Please take a look at
https://en.wikipedia.org/wiki/Mechanical_advantage

Regards,
Buzz

Thanks. I'll see if I can apply it here.

 

[Priyadarshini]

No, both rod and disc rotate about an axis through their centres.
I suppose you could consider that the not-yet-wound string instantaneously rotates about the point of contact with the disc, but since it exerts no torque about that point it is quite irrelevant.

How do you know that the string at the contact point of the disc doesn't exert any torque?

 

[haruspex]

How do you know that the string at the contact point of the disc doesn't exert any torque?

I meant its rotation about the contact point exerts no torque. It is a flexible string, not a rigid lever. Only the tension in the string exerts a torque, but that has no moment about the contact point: torque = force x perpendicular distance, and the distance from the string to the contact point is zero.
The torque of interest exerted by the tension is at the disc's axis.

 

[physics user1]

The string exerts a torque at the contact point of the disk (F*R), the torque is because of the 900 N force, to balance this torque you need the torque of the lever that is needed to be minimum equal to the torque at the disk

 

[Priyadarshini]

The string exerts a torque at the contact point of the disk, the torque is because of the 900 N force, to balance this torque you need the torque of the lever that is needed to be minimum equal to the torque at the disk

I meant its rotation about the contact point exerts no torque. It is a flexible string, not a rigid lever. Only the tension in the string exerts a torque, but that has no moment about the contact point: torque = force x perpendicular distance, and the distance from the string to the contact point is zero.
The torque of interest exerted by the tension is at the disc's axis.

Oh, I see! Thank you. It works now.

 

[CWatters]

Isn't that 0.4*900?

0.2 * 900

 

[Priyadarshini]

0.2 * 900

Thanks.