Solving a Pulley System: Friction, Tension & Acceleration

AI Thread Summary
The discussion focuses on solving a pulley system problem involving friction, tension, and acceleration. The user attempts to apply Newton's second law by summing forces for each block but struggles with the correct signs in their equations. Key advice includes drawing free body diagrams for each block and ensuring the correct relationship between tension and acceleration. The user realizes that the weight of block C must exceed the tension for it to accelerate downward, prompting a correction in their calculations. Ultimately, understanding the direction of forces is crucial for arriving at the correct answer.
bungie77
Messages
12
Reaction score
0

Homework Statement



http://img214.imageshack.us/img214/6531/pulleysystemph6.th.jpg

Homework Equations



sum of forces

The Attempt at a Solution



So I tried summing the forces in the x-direction where

m*a (block a) = -Ff + T = m(block a)*g*u + m*a(block b) + m*g(block b)

where u is the coefficient of friction, T is the tension in the rope and Ff is the frictional force.

I'm not, however, getting an answer on the multi choice list. Is there something I'm missing?
 
Last edited by a moderator:
Physics news on Phys.org
You've got to look at each block separately when doing these problems. This is essential. Draw a free body diagram for each. Identify the forces acting on each block and use Newton 2. You'll have 2 equations with 2 unknowns that you can solve. Note that the magnitudes of the tension and acceleration for each block are the same.
 
well I've been doing that but i can't seem to get the right answer... i was wondering if maybe i was screwing something up with the directions...

so far I've come to the following;

for block A;

Fx = T - Uk*N = mA*ax
Fy = N - mA*g = 0

So essentially, Fx = T - Uk*mA*g = mA*ax

For block C;

Fy = T - mC*g = mC*ay

I understand ay = ax but i can't seem to get the right answer... I'm getting something around 1.6 :(
 
bungie77 said:
well I've been doing that but i can't seem to get the right answer... i was wondering if maybe i was screwing something up with the directions...

so far I've come to the following;

for block A;

Fx = T - Uk*N = mA*ax
Fy = N - mA*g = 0

So essentially, Fx = T - Uk*mA*g = mA*ax

For block C;

Fy = T - mC*g = mC*ay

I understand ay = ax but i can't seem to get the right answer... I'm getting something around 1.6 :(
Oh, good, you are on track; only error you made was for block C, since it accelerates down, its weight must be greater than T, so your plus/minus signs are incorrect. It should be mCg - T = mC*ay. Do you see why?
 
ahhh yeah of course. and since the tension is obviously greater in block A the signs are opposite to block C. thanks :)
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top