Pulley

prosoccer
Posers and Puzzles 02 Apr '07 22:40
1. 02 Apr '07 22:40
A 75kg mass is attached to one end of a rope and a 350kg mass is attached to the other end. The rope is strung over a single pulley so that the 350kg mass will begin to accelerate downward (obviously) and at the same time pull the 75kg mass up. The coefficient of friction of the pulley and rope is 0.179. What acceleration does the system achieve (assuming no air resistance) and what is the tension in the rope?
2. 03 Apr '07 00:30
Originally posted by prosoccer
A 75kg mass is attached to one end of a rope and a 350kg mass is attached to the other end. The rope is strung over a single pulley so that the 350kg mass will begin to accelerate downward (obviously) and at the same time pull the 75kg mass up. The coefficient of friction of the pulley and rope is 0.179. What acceleration does the system achieve (assuming no air resistance) and what is the tension in the rope?
Cancel out that friction in the pulley.
3. 03 Apr '07 02:03
Originally posted by prosoccer
A 75kg mass is attached to one end of a rope and a 350kg mass is attached to the other end. The rope is strung over a single pulley so that the 350kg mass will begin to accelerate downward (obviously) and at the same time pull the 75kg mass up. The coefficient of friction of the pulley and rope is 0.179. What acceleration does the system achieve (assuming no air resistance) and what is the tension in the rope?
Downward force = (350-75)*9.8 = 2695 N.
Frictional force (I think) = 0.179 * (350+75) * 9.8 = 745.5 N

Net downward force = 1950 N
Total mass = 425 kg

Acceleration = 1950/425 = 4.59 m/s^2.

Sounds reasonable, but I take no responsibility for the accuracy of the above results. ðŸ˜€
4. 03 Apr '07 12:39
Originally posted by GregM
Downward force = (350-75)*9.8 = 2695 N.
Frictional force (I think) = 0.179 * (350+75) * 9.8 = 745.5 N

Net downward force = 1950 N
Total mass = 425 kg

Acceleration = 1950/425 = 4.59 m/s^2.

Sounds reasonable, but I take no responsibility for the accuracy of the above results. ðŸ˜€
I got a ~6 m/s^2 but that's because I didn't factor in friction because I wasn't sure how to in a pulley like this.
5. PBE6
Bananarama
03 Apr '07 13:28
Originally posted by prosoccer
I got a ~6 m/s^2 but that's because I didn't factor in friction because I wasn't sure how to in a pulley like this.
All the pulley does is redirect the force, so you can simplify this problem by thinking of it as two blocks sitting on a table, tied together, and being tugged from either end. The pulley friction is now analogous to the friction of the blocks moving against the table.
6. sonhouse
Fast and Curious
04 Apr '07 14:29
Originally posted by PBE6
All the pulley does is redirect the force, so you can simplify this problem by thinking of it as two blocks sitting on a table, tied together, and being tugged from either end. The pulley friction is now analogous to the friction of the blocks moving against the table.
The only problem I have with the friction co-efficient is this:
I am not an expert in this so correct me please if I am wrong.
The co-efficient is a dimensionless number, in this case, 0.18 or so.
Now if I understand friction, wouldn't it have a definite imparement value, imparement of movement in a particular dimension?
That is to say, suppose you have a brick on a concrete floor and a string tied to it and you are moving the brick parallel to the floor. Now under these circumstances, the brick takes a certain force to start it moving, but isn't it an actual number like 100 grams for that brick.
If you double the friction, then it takes 200 grams of force to move.
I guess that makes it a ratio between the force and the weight/mass.
But what about pulling towards different dimensions? Doesn't that introduce a vector to it that modifies the original friction co-efficient?
If I now pull the brick upwards at a 45 degree angle with respect to the floor, isn't the friction now less by a certain amount?
7. PBE6
Bananarama
04 Apr '07 15:43
Originally posted by sonhouse
The only problem I have with the friction co-efficient is this:
I am not an expert in this so correct me please if I am wrong.
The co-efficient is a dimensionless number, in this case, 0.18 or so.
Now if I understand friction, wouldn't it have a definite imparement value, imparement of movement in a particular dimension?
That is to say, suppose you have ...[text shortened]... 45 degree angle with respect to the floor, isn't the friction now less by a certain amount?
The coefficient of friction is the ratio of the frictional force (force resisting movement) to the normal force (the force squeezing the two surfaces together). If you pull a block along a table using a string, and the string is angled upwards, part of the force will be directed upwards effectively pulling the two surfaces apart. This will result in less friction being generated.
8. sonhouse
Fast and Curious
04 Apr '07 18:132 edits
Originally posted by PBE6
The coefficient of friction is the ratio of the frictional force (force resisting movement) to the normal force (the force squeezing the two surfaces together). If you pull a block along a table using a string, and the string is angled upwards, part of the force will be directed upwards effectively pulling the two surfaces apart. This will result in less friction being generated.
So there would be a polar co-ordinate vector added to the prime friction number?
BTW, wouldn't you have to use 1 minus the co-efficent in your calcs? You went 0.179X (g forces). shouldn't it be 1-.179 or 0.82 times the resultant accel? If I have it right, the 0.179 is the stopping force and if it totally stopped the motion, it would by definition be 1.0 and if it had zero effect on the motion it would be 0.0, right?
So if it had a 0.179 stopping value it seems the total force would be 1-0.179 or 0.821 which would mean the actual acel would be 82% of what it would be without friction, right?
9. PBE6
Bananarama
04 Apr '07 18:431 edit
Originally posted by sonhouse
So there would be a polar co-ordinate vector added to the prime friction number?
BTW, wouldn't you have to use 1 minus the co-efficent in your calcs? You went 0.179X (g forces). shouldn't it be 1-.179 or 0.82 times the resultant accel? If I have it right, the 0.179 is the stopping force and if it totally stopped the motion, it would by definition be 1.0 a ...[text shortened]... 821 which would mean the actual acel would be 82% of what it would be without friction, right?
I didn't do the calculation, GregM did. But I would do it like this.

Imagine the blocks are lying flat on a table, connected by a rope, with forces equal to the gravitational forces in the original question being applied to each block. The 350 kg is on the right. Let right be +. Therefore the sum of the forces on the blocks is as follows:

sum(F) = F1 - F2 - Ff

F1 = 350 kg * 9.81 N/kg = 3434 N
F2 = 75 kg * 9.81 N/kg = 736 N
Ff = 0.179 * (350 kg + 75 kg) * 9.81 N/kg = 746 N

sum(F) = 3434 - 736 - 746 = 1952 N

sum(F) = m * a --> a = sum(F)/m

a = 1952 N / (350 kg + 75 kg) = 4.59 m/s^2

Therefore the acceleration will be 4.6 m/s^2 to the right. Bringing this back into the original model, assuming the 75 kg weight is on the left and the 350 kg weight is on the right, the acceleration will be 4.6 m/s^2 in the clockwise direction about the pulley centre.

Without friction, our equation is as follows:

sum(F) = F1 - F2 = 3434 N - 736 N = 2698 N

a = sum(F)/m = 2698 N / (350 kg + 75 kg) = 6.35 m/s^2

Therefore, using the same convention noted above, the acceleration will be 6.4 m/s^2 in the clockwise direction. Taking the ratio of the two answers, we find that:

a(friction)/a(no friction) = (4.59 m/s^2) / (6.35 m/s^2) = 0.723

Therefore the friction reduces the acceleration to 72% of the acceleration in the frictionless system.
10. sonhouse
Fast and Curious
04 Apr '07 22:08
Originally posted by PBE6
I didn't do the calculation, GregM did. But I would do it like this.

Imagine the blocks are lying flat on a table, connected by a rope, with forces equal to the gravitational forces in the original question being applied to each block. The 350 kg is on the right. Let right be +. Therefore the sum of the forces on the blocks is as follows:

sum(F) = F1 ...[text shortened]... re the friction reduces the acceleration to 72% of the acceleration in the frictionless system.
Wouldn't that work out to 1-0.723 = 0.277 as the friction co-effiecient?
In the original statement the co# was 0.179. What's with this discrepency?
11. 05 Apr '07 00:02
Is this another 'homework' time?
12. sonhouse
Fast and Curious
05 Apr '07 13:30
Originally posted by Gastel
Is this another 'homework' time?
You think?ðŸ™‚
13. PBE6
Bananarama
05 Apr '07 14:34
Originally posted by sonhouse
Wouldn't that work out to 1-0.723 = 0.277 as the friction co-effiecient?
In the original statement the co# was 0.179. What's with this discrepency?
If you work out the problem algebraically, the final result is:

a = g*(m1-m2)/(m1+m2) - g*mu

where:

"g" is the acceleration due to gravity
"m1", "m2" are the masses of the blocks
"mu" is the coefficient of kinetic friction

Therefore you'd expect the discrepancy to be g*mu, not just mu. This works out to 9.81 * 0.179 = 1.76 m/s^2, which is equal to the difference between 6.35 m/s^2 and 4.59 m/s^2.
14. sonhouse
Fast and Curious
05 Apr '07 17:53
Originally posted by PBE6
If you work out the problem algebraically, the final result is:

a = g*(m1-m2)/(m1+m2) - g*mu

where:

"g" is the acceleration due to gravity
"m1", "m2" are the masses of the blocks
"mu" is the coefficient of kinetic friction

Therefore you'd expect the discrepancy to be g*mu, not just mu. This works out to 9.81 * 0.179 = 1.76 m/s^2, which is equal to the difference between 6.35 m/s^2 and 4.59 m/s^2.
Ok, you subtracted out the friction not multiplied. I thought you had originally multiplied by 0.179. I see what you did now.