If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Class 11 Physics (India)>Unit 11

Lesson 4: Torque and equilibrium

# Moments (part 2)

2 more moment problems. Created by Sal Khan.

## Want to join the conversation?

• Why aren't the clockwise forces marked with a negative sign when Sal does the algebra? As always, thanks a bunch.
• The reason why there is no negative sign in the equation is because He is setting it as in equilibrium, the rightwise is equal to the counter-clockwise. This is like up is equal to down, it is implied that the forces are already acting in opposite directions so the negative sign is not needed. Another way to look at this would be The net movement of the table equals zero; 0 = clockwise force minus counter-clockwise force. Add counter-clockwise force on each side and you get. clockwise force = counter-clockwise force.
• I'm not sure whether or not I may have missed a title, but is there a video that addresses the parallel axis theorem? I have a textbook that briefly addresses the theorem, but it gives a poor example and moves right along.
• I don't think there is one specifically for that topic, but the parallel axis theorem as I know states that the moment of inertia of a body about any axis is equal to the sum of [the moment of inertia of the body about an axis (parallel to the axis taken) at its center of mass] and [the product of the mass of the body and the square of the distance between the two parallel axes].
That is : I = Icm + (m).(d^2)
{where I is the moment of inertia about the axis taken; Icm is the moment of inertia about the center of mass and parallel to the axis taken; And m is the mass of the body, d is the distance between the two parallel lines}

I hope the statement wasn't so confusing...
• I didn't get how the forces of 20 N and and the force exerted by the book are clockwise ?
• we considered the point where the left leg meets the top of the table to be the pivot point (or the point around which the table rotate). With respect to that point, the 20N force and the force exerted by the book are clockwise.
• please can give me answer of this ques , 10N weight can be lifted in one end of a thin rod by applying a 5N force on the
other end. If the length of the rod is 1.5 meter, where is the fulcrum situated? and please tell me how to measure
• The fulcrum is located somewhere between the two weights. The distance from the fulcrum for the 5N weight is twice the distance from the fulcrum (on the opposite side) for the 10N weight.
• when calculating the counterclockwise forces, why isn't the weight of the right leg counted? only the force up is counted not the one going down? why is that ?
• I'm 12 years old and I understand this. My friends says that it is impossible and my dad says if I learn to much I might get really stressed. I don't think learning stresses me out much. Should I continue? I really like learning physics.
• If you are comfartable and you enjoy it. Then why not! However listen to your parents too. I mean try to find an optimized way. Do extra but don't burn your head. It's good that you learn more but everything has a time. Just don't push so hard.
• Wouldn't the axis of rotation be at the base of the left leg? If you removed the right leg, the table would rotate on the base of the left leg. If the axis was at the top, that left leg would slip off the floor and rotate clockwise. That would be weird.
• Yes you are right...the axis should be at the base of the left leg...but when calculating the torque it wont matter as perpendicular distance of the 100 N force from the axis will be same in both cases(you know, since force is a vector you can translate the force vector or extrapolate its line of action downwards without changing its value) ....

Hope this helps :)