Register to reply

Forces and Pulley System

by Soaring Crane
Tags: forces, pulley
Share this thread:
Soaring Crane
#1
Sep19-06, 06:25 AM
P: 484
Block A has weight w_A and block B has weight w_B. Block A is positioned on the horizontal surface of a table. Block A is connected to a cord passing over an easily turned pulley that has Block B hanging down from the pulley. Once Block B is now set into downward motion, it descends at a constant speed.

a. Calculate the coefficient of kinetic friction between Block A and the tabletop.

I drew Block A's force diagram. For this part, I am assuming Block A is not in motion as yet; it is just sitting on the tabletop. (Is this correct?)

Well, F_fr = mu_K * F_N and the normal force = w_A.

F_fr = mu_k*w_A

mu_k = F_fr/w_A

But I think this is somewhat incorrect. Musn't I express mu_k with the known variables give. What is the Value for F_fr?

b. A cat, also of weight w_A, falls asleep on top of block A. If Block B is now set into downward motion, what is its acceleration (magnitude and direction)? Express a in w_A, w_B, and g.

Well, I am unsure of the F_fr expression. This is what I did for this part.

For Block A, the expression is now F_T - F_fr = [(2*w_A)/g]*a
For Block B, ---------------------w_B - F_T = (w_B/g)*a

Adding these two expressions, I get

w_B - F_fr = [(2*w_A)/g]*a + (w_B/g)*a
w_B - F_fr = a[(2*w_A + w_B)/g]

a = [w_B - F_fr]/[(2*w_A + w_B)/g]
a = g*[(w_B-F_fr)/(2*w_A + w_B)]

But I don't know what to put for F_fr as noted in the first part.

Are my reasonings and math that I did do correct?

Thanks.
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
Astronuc
#2
Sep19-06, 07:08 AM
Admin
Astronuc's Avatar
P: 21,827
Block A has weight w_A and block B has weight w_B. Block A is positioned on the horizontal surface of a table. Block A is connected to a cord passing over an easily turned pulley that has Block B hanging down from the pulley. Once Block B is now set into downward motion, it descends at a constant speed.

a. Calculate the coefficient of kinetic friction between Block A and the tabletop.

I drew Block A's force diagram. For this part, I am assuming Block A is not in motion as yet; it is just sitting on the tabletop. (Is this correct?)

Well, F_fr = mu_K * F_N and the normal force = w_A.

F_fr = mu_k*w_A

mu_k = F_fr/w_A

But I think this is somewhat incorrect. Musn't I express mu_k with the known variables give. What is the Value for F_fr?
One may assume block A has zero velocity.

To find F_fr, what is F_fr opposing? What is the significance of Block B descending at constant speed?
Soaring Crane
#3
Sep20-06, 06:07 AM
P: 484
F_fr = w_B, but when I plug this value into the expression for a that I found, acceleration would be 0. According to Newton's First Law, a body that has constant velocity and, therefore, 0 acceleration is acted on by no net forces. Where do I go from here?

wxrocks
#4
Sep20-06, 07:23 AM
P: 131
Forces and Pulley System

Soaring -- didn't your last post just answer your own question??? In your first post, you wanted to know what F_fr was. So, what is it?


Register to reply

Related Discussions
Forces in a Pulley System Introductory Physics Homework 3
Pulley system Introductory Physics Homework 2
Pulley system Introductory Physics Homework 3
Forces involved in spring-pulley system Introductory Physics Homework 6
Two pulley system Introductory Physics Homework 14