Attempt: How to Calculate Acceleration in Pulley Systems

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Homework Help Overview

The discussion revolves around calculating acceleration in pulley systems, specifically involving multiple masses, tension, and static equilibrium. The original poster presents several problems related to forces acting on masses connected by strings over pulleys, including scenarios with static and kinetic friction.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply concepts of static equilibrium and tension to the problems but expresses uncertainty about how to approach the calculations. Some participants question the clarity of the diagrams provided and emphasize the need for the original poster to show their work. Others suggest considering the forces acting on the masses and how they relate to acceleration.

Discussion Status

Participants are engaging with the problems, with some providing insights into the nature of the forces involved. There is a recognition that calculations are necessary for the second problem, and guidance has been offered to help the original poster think through the forces acting on the system.

Contextual Notes

There is a mention of the original poster's lack of experience with inclined planes, which may affect their ability to tackle the third problem. Additionally, the requirement for participants to demonstrate their attempts before receiving help is noted.

starprincess
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Hello Everyone,
I am having difficulty with these selected problems. Attached is a link to the diagram in reference to each problem. I would truly appreciate your help! Thank you!

1. In Fig.1 the scale at the left is attached to the ceiling and a mass of 1.00 kg hangs from it. It reads 9.81 N. The identical scale at the right is connected by perfect strings passing over perfect pulleys to two 1.00 kg masses hanging vertically at the end of the strings. The scale at the right reads:
a) exactly 9.81 N
b) more than 9.81 N, but not quite as much.
c) less than 9.81 N.
d) exactly 19.81 N.
e) more than 19.62 N.

http://i82.photobucket.com/albums/j246/starprincess101/DSC01143.jpg"

Attempt: I believe this has something to do with static equilibrium equation and the use of tension? So I am guessing it would be A.2. Two identical masses are attached by a light string that passes over a small pulley, as shown in Fig.2. The table and the pulley are frictionless. The masses are moving:
a) with an acceleration less than g.
b) at constant speed.
c) with an acceleration greater than g.
d) with an acceleration equal to g.

Attempt: I think it would be constant speed.http://i82.photobucket.com/albums/j246/starprincess101/DSC01144.jpg" 3. Two masses are connected by a string which goes over an ideal pulley as shown in Fig. 3. Block A has a mass of 3.00 kg and can slide along a rough plane inclined 30.0 degrees to the horizontal. The coefficient of the kinetic friction between block A and the plane is 0.400. Block B has a mass of 2.77 kg. What is the acceleration of the blocks?
a) 0.392 m/s2
b) 1.96 m/s2
c) 3.12 m/s2
d) 5.35 m/s2
e) 0

Attempt: I have no idea how to solve this problem or where to start because I have never learned about inclinations yet.

http://i82.photobucket.com/albums/j246/starprincess101/DSC01145-1.jpg"
 
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Your diagrams are far too small to read!
Also, you must show your attempt to do the problems before we can help you.
 
Fixed pictures!
 
In problem one, the two scales are doing exactly the same job. The weight on the left pulley is holding the scale while the weight on the right pulls on it with 9.81 N. You might think it would read 2x9.81 in both cases because there is 9.81 pulling on each end, but the scale is designed to read just 9.81 N.

For problem 2, you must do some calculations. Think of the two masses and the string as one object that may have an acceleration to the right and down (always in the direction of the string).
The sum of the forces in the direction of the string must equal ma.
Have a go at it - add up all the forces acting that way and put " = ma" on the end!
Show us what you get and we'll check it out for you.
 

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