Mind boggled by a tension problem.

In summary: The vertical forces are the tension in the rope pulling up on the clown's hands, the friction pulling to the right on his feet, and gravity pulling down on him.The horizontal forces are the tension in the rope pulling to the left on his feet, and the friction pulling to the right on his feet.The net force on the clown is the sum of the vertical and horizontal forces: 3 + 2 = 5.
  • #1
bananan
176
0
[img=http://img285.imageshack.us/img285/2332/untitled2bj9.th.jpg]

Hi everyone! I am mind boggled by this tension problem.
Here's what I grasp. If the clown pulls downward, the rope exerts an opposite force with the same magnitude upward. I don't understand how exactly the tension going upward exerts a tension along the three pulleys. Can anyone draw a free body diagram of the tensions throughout the pulleys and do his feet? Everytime I do it, the tension from last pulley points toward his feet, and that doesn't make any sense of course, since his feet is getting pulled backwards. I'm sure once I get the concept of tension I can finish the problem. Thanks for checking this out!
 
Physics news on Phys.org
  • #2
Brr, a clown [:yuck:]. I refuse to thing about anything. [:smile:]
 
  • #3
Try drawing the forces on his arms and the force on his legs only.
 
  • #4
bananan said:
Here's what I grasp. If the clown pulls downward, the rope exerts an opposite force with the same magnitude upward.
Absolutely correct.
I don't understand how exactly the tension going upward exerts a tension along the three pulleys.
Assuming the rope is massless and the pulleys frictionless, the tension created in the rope will be the same throughout the rope. If the clown pulls down on the rope with a force of X Newton's, that means the tension force exerted by the rope (at both ends) is X Newtons: The rope pulls up on the clown (at one end) and to the left on his feet (at the other end) with the same force.
Can anyone draw a free body diagram of the tensions throughout the pulleys and do his feet?
The tension is the same throughout the rope.
Everytime I do it, the tension from last pulley points toward his feet, and that doesn't make any sense of course, since his feet is getting pulled backwards.
Why doesn't that make any sense?
 
  • #5
well, if the clown is pulling downward on the rope, and the tension is the reaction of the clown's action. Then the clown is stationary since the forces are at equilibrium right? But while all this is happening, the clown's feet is touching the floor, where there is friction which in turn opposes the clown's action. So I don't understand how can there be two forces (the tension of the rope and the friction) opposing one force (the pulling down)? I thought for every force applied, there has to be an opposing force to it. I guess what I'm confused at then is how all the forces work in the problem.
 
  • #6
bananan said:
well, if the clown is pulling downward on the rope, and the tension is the reaction of the clown's action.
The clown pulling down on the rope, and the rope pulling up on the clown are "action-reaction" pairs, yes.

Then the clown is stationary since the forces are at equilibrium right?
Careful: Action-reaction pairs never produce equilibrium--they act on different objects. (This is important to understand.)

The clown remains stationary as long as the net force on the clown is zero.

But while all this is happening, the clown's feet is touching the floor, where there is friction which in turn opposes the clown's action.
What horizontal forces act on the clown? I count two: The rope pulling to the left and friction pulling to the right. As long as they balance, the clown will not be pulled off his feet.

So I don't understand how can there be two forces (the tension of the rope and the friction) opposing one force (the pulling down)? I thought for every force applied, there has to be an opposing force to it.
The rope exerts two different forces on the clown: It pulls up on the clown's hands (reaction: the clown's hands pull down on the rope) and it also pulls sideways on the clown's feet (reaction: the clown's feet pull back on the rope). These two forces happen to have the same magnitude, since they are both produced by the tension in the rope, but they are different forces.

The friction is another force acting on the clown; it's a force of the ground pulling on the clown's feet--the reaction to the friction is the clown's feet pulling back on the ground.

To find out what happens to the clown, analyze the forces acting on the clown.

What forces act on the clown? (Consider vertical and horizontal forces separately.) I count 3 vertical forces and 2 horizontal forces acting on the clown.
 

What is a tension problem?

A tension problem refers to a situation where there is a force acting on an object, causing it to stretch or deform. This force can be internal or external and can be calculated using various equations from the laws of physics.

How do you solve a tension problem?

To solve a tension problem, you need to first identify all the forces acting on the object and their directions. Then, you can use equations such as Newton's second law of motion or Hooke's law to calculate the tension force. It is important to use consistent units and check your work for accuracy.

What are some real-life examples of tension problems?

Some real-life examples of tension problems include a rope pulling on a box, a bridge supporting the weight of cars and trucks, and a bungee jumper being pulled by the elastic cord. Tension problems can also occur in muscles, tendons, and ligaments in the human body.

What are the common mistakes when solving a tension problem?

One common mistake when solving a tension problem is forgetting to include all the forces acting on the object. Another mistake is using the wrong equation or using inconsistent units. It is also important to pay attention to the direction of forces and use the correct signs in calculations.

Why is understanding tension problems important in science?

Understanding tension problems is important in science because it helps us understand the behavior of objects under different forces. This knowledge is crucial in fields such as engineering, physics, and biomechanics. It also allows us to design and build structures that can withstand different levels of tension to ensure safety and efficiency.

Similar threads

  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
772
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Replies
37
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Back
Top