Help with Torque and Force Please

Also, the child's weight is not acting perpendicular to the see-saw at the pivot point, so you need to consider the component of the weight vector that is perpendicular to the see-saw in your calculation.
  • #1
o0ojeneeo0o
5
0

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
the see-saw? Give a positive answer. Answer
in units of N.

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


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
  • #2
Welcome to PF.

How would you think to start?
 
  • #3
Just a little bit short on work on your part! Please make an effort first.
 
  • #4
Well, try to look up for equations for Torque.Its a pretty straightforward problem
 
  • #5
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!
 
  • #6
1. [tex]\Sigma[/tex][tex]\tau[/tex] = -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?
 
  • #7
o0ojeneeo0o said:
1. [tex]\Sigma[/tex][tex]\tau[/tex] = -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.
 

Related to Help with Torque and Force Please

1. What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance between the axis of rotation and the point where the force is applied.

2. How is torque different from force?

Torque and force are related but different concepts. While force is a push or pull on an object, torque is the measure of the turning force that causes an object to rotate. Force is a vector quantity, meaning it has both magnitude and direction, while torque is a rotational vector, meaning it has a direction of rotation.

3. What are some real-life examples of torque?

Some common examples of torque include opening a door, using a wrench to tighten a bolt, or twisting the lid off a jar. In each of these scenarios, a force is applied at a distance from the axis of rotation, creating a torque that causes the object to rotate.

4. How is torque related to angular acceleration?

Torque and angular acceleration are directly proportional, meaning as one increases, the other also increases. This relationship is described by the equation T = Iα, where T is torque, I is the moment of inertia, and α is the angular acceleration.

5. How can I calculate torque?

To calculate torque, you will need to know the force applied to an object and the distance between the axis of rotation and the point where the force is applied. The formula for torque is T = F x d, where T is torque, F is force, and d is distance. Torque is typically measured in units of newton-meters (Nm) or foot-pounds (ft-lb).

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
225
Replies
6
Views
904
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
392
  • Introductory Physics Homework Help
Replies
8
Views
551
  • Introductory Physics Homework Help
Replies
11
Views
924
  • Introductory Physics Homework Help
Replies
7
Views
915
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
819
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top