Homework Help: Help with Newton's law

1. Oct 25, 2004

newcool

Hi, I have been given this problem to do as extra credit:

An inventive child named Pat wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley, Pat pulls on the loose end of the rope with such a force that the spring scale reads 250 N. Pat’s true weight is 320 N, and the chair weights 160 N.

Show that the acceleration of the system of Pat and the chair is upward and find its magnitude.

Find the force Pat exerts on the chair.

Diagram:
http://www.geocities.com/newc00l/physics.html

I made 3 free body diagrams, one of the child with the chair, one of the spring scale and one of the spring inside the spring scale, but am at a loss to generate the necessary equations and am stuck.

Thanks in advance for any help.

2. Oct 25, 2004

spacetime

The scale tells you the tension in the rope. The rope pulls on the child with the same force as the child pulls on the rope - which is the same as tension.
Moreover, the chair is also pulled by the rope with the same force. So the total upward force on the child is twice of the tension in the rope.

The total downward force is known. So, the answer is easily calculated.

spacetime
www.geocities.com/physics_all/index.html

3. Nov 2, 2004

newcool

I made a free body diagram and solved part A to be .408 m/s/s
500
^
|
Pat and Chair ^A
49 kg
|
480 N

500 - 480 = 49a
20 = 49a
a = .408

How do you solve part B?

4. Nov 2, 2004

Leong

I think when Pat pulls the loose end of the rope, the downward forces acting on the scale will be the pulling force of 250 N plus the weigh of Pat and the chair. If the scale is consider to be massless, the from Newton's 2nd Law :
T= 250 N + 320 N +160N=730N

5. Nov 2, 2004

newcool

Hey Leong,

if Pat exerts 730 newtons on the chair then the whole system would fly down.

The answer is somewhere between 50 and 100 newtons.

6. Nov 2, 2004

Leong

Pat doesn't exert 730 N, he only exerts 250 N the rest is coming from his and the chair weight.

7. Nov 2, 2004

Leong

If what i assume is correct, then the accleration will be 5.11 m/s^2 and the normal force Pat exerts on the chair is 487 N.

8. Nov 2, 2004

newcool

The tension in the rope is 2T = 250 *2 = 500 because it's a pulley system. I think you made a mistake somewhere because the answer is between 50 and 100 newtons.

9. Nov 2, 2004

Leong

Sorry if i have misguided you.

10. Nov 2, 2004

newcool

no problem, I thank you for your efforts because this is not an easy problem. That's the whole point of these forums, to get people to try to help others on hard problems. Trying is the important part.

11. Nov 4, 2004

Leong

I went to a bookstore today to buy a book but it wasn't there anymore. because i was there already, i went through a few physics book to find a similar problem to yours and i found exactly the same question in one book and it had the answers. a=0.409 m/s^2 and N=83.3 N.
consider the chair : there are 3 forces acting on the chair. The normal force, N, Pat exerts on the chair, its weight and the tension of the rope. so, from Newton's 2nd Law :
$$T-W_{chair}-N=m_{chair}a$$
then, N=83.3 N

Last edited: Nov 4, 2004
12. Nov 4, 2004

HallsofIvy

Pat weighs 320 N but since he/she is pulling down on the rope with 250 N the rope is pulling up on him with 250 N. He is "pressing down" on the chair with 320-250= 70 N. The total weight he has to lift is 70+ 160= 230 N. Since that's less than 250 N, there is a "lifting" force of 250-230= 20 N.