# Help with Torque and Force Please!

## Homework Statement

A child weighing 38 N climbs onto the right
end of a see-saw, little knowing that it is tied
down with a rope at its left end. The see-saw
is 2.6 m long, weighs 12 N and is tilted at an
angle of 17◦ from the horizontal. The center
of mass of the see-saw is half way along its
length, and lies right above the pivot.

1. What torque does the weight of the child
exert about the pivot point? Take counter-
clockwise to be positive. Answer in units of
Nm.

2. What torque does the weight of the see-saw
exert about the pivot point? Take counter-
clockwise to be positive. Answer in units of
Nm.

3. What torque does the rope exert about the
pivot point? Take counter-clockwise to be
positive. Answer in units of Nm.

4. What force does the rope exert downwards on
in units of N.

5. What is the total positive force exerted up-
wards by the pivot? Answer in units of N.

## The Attempt at a Solution

LowlyPion
Homework Helper
Welcome to PF.

How would you think to start?

Just a little bit short on work on your part! Please make an effort first.

Well, try to look up for equations for Torque.Its a pretty straightforward problem

Umm, I know it would probably seem like it would be straight forward, but I'm a complete idiot when it comes to physics. I also contribute that in part to the fact that my physics professor does not speak English very well and doesn't answer any of our questions during class so I've been trying to teach myself. That hasn't gone very well. Sorry. I'll post my attempt shortly... thank you!!!

1. $$\Sigma$$$$\tau$$ = -mg - Fchild
= -12N(.65m) - 38N(1.3m)
= -57.2Nm

I don't think that's right though... the answer shouldn't be negative should it?

LowlyPion
Homework Helper
1. $$\Sigma$$$$\tau$$ = -mg - Fchild
= -12N(.65m) - 38N(1.3m)
= -57.2Nm

I don't think that's right though... the answer shouldn't be negative should it?

When you take the sum of the Torques you take the moments acting about the pivot point. Since the see-saw itself has a center of mass at the pivot point it can't contribute to the net torque.

The child does act at a distance along the see-saw of 1.3m, but you have failed to account for the 17 degree tilt. Since that moves the child effectively closer to the pivot, then the effective moment arm for the child's mass is incorrect.