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Horizontal force due to tension from vertical wieght

by fatalphysics
Tags: box, force, physics help, pulley
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fatalphysics
#1
Oct18-11, 04:25 PM
P: 1
1. The problem statement, all variables and given/known data

Block C (m= 4 kg) sits on a frictionless horizontal surface. Block B of (m
= 2kg) sits on top of block C, and is attached to a rope that runs over the massless
pulley as shown in the figure. Block A (m=1 kg) hangs vertically from the rope. With
what horizontal force F must you push Block C so that block A rises with an upward
acceleration of a = 3m/s2? All surfaces are frictionless.

Diagram:
Click image for larger version

Name:	question7pic.png
Views:	0
Size:	11.9 KB
ID:	40143


2. Relevant equations

F=ma


3. The attempt at a solution

Using free body diagrams came up with:

F=mac
T=2ac
T/2=ac

F=maa
T-mg=(3)(1)
T=3+(9.8)(1)
T=12.8 N

(12.8)/2=ac
ac=6.4 m/s2

F=(mc+ma)ac
F=(6)(6.4)
F=38.4 N

Actual answer = 59.8N
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PeterO
#2
Oct18-11, 07:28 PM
HW Helper
P: 2,318
Quote Quote by fatalphysics View Post
1. The problem statement, all variables and given/known data

Block C (m= 4 kg) sits on a frictionless horizontal surface. Block B of (m
= 2kg) sits on top of block C, and is attached to a rope that runs over the massless
pulley as shown in the figure. Block A (m=1 kg) hangs vertically from the rope. With
what horizontal force F must you push Block C so that block A rises with an upward
acceleration of a = 3m/s2? All surfaces are frictionless.

Diagram:
Attachment 40143


2. Relevant equations

F=ma


3. The attempt at a solution

Using free body diagrams came up with:

F=mac
T=2ac
T/2=ac

F=maa
T-mg=(3)(1)
T=3+(9.8)(1)
T=12.8 N

(12.8)/2=ac
ac=6.4 m/s2

F=(mc+ma)ac
F=(6)(6.4)
F=38.4 N

Actual answer = 59.8N
It looks like you are mixing up the mass of A and B. A is 1 ; B is 2 not the other way.


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