- #1
allstar_jason
- 2
- 0
Hi, I'm trying to work through a pulley problem in Hibbeler's Statics and Dynamics (11th edition). I'm working on 13-26 for those who have the text. The problem in involves a block, A of weight 100-lbs moving down an incline at 5 ft/s. It's attached to a pulley system with the first pulley stable on the roof and the rope goes through another pulley, C, below pulley A and up to the roof. Directly attached to pulley C is pulley D which has a cord running through it. One end of the cord is attached to the ground, the other has a mass, B, of weight 50-lbs. The coefficient of kinetic friction is 0.2. Here's a rough sketch, A is attached at O and the slope A is on has an angle of 36.9 degrees:
O |
| |
| |
| |
C
D
| |
| |
B |
Anyways, the problem wants to determine the acceleration of A and the distance A travels before stopping, and mass of the pulleys and cables can be neglected.
So for A, it should be [tex]\sum[/tex]Fx = 100 * sin(36.9) - N * 0.2 - T = massA*accelA
where N is just the normal force: 100*cos(36.9)
And I think the force's on B would be [tex]\sum[/tex]Fy = 50 - T = massB*accelB
Now, that leaves the lengths of cord, which should give two equations as it's two separate cords. I believe the length of the cord directly attached to A will be Sa + 2Sc = L1
But I'm unsure of the second cord so here's where I'm stuck. Also I'm not entirely sure everything else I've done is right either so if someone wouldn't mind checking that would be great.
O |
| |
| |
| |
C
D
| |
| |
B |
Anyways, the problem wants to determine the acceleration of A and the distance A travels before stopping, and mass of the pulleys and cables can be neglected.
So for A, it should be [tex]\sum[/tex]Fx = 100 * sin(36.9) - N * 0.2 - T = massA*accelA
where N is just the normal force: 100*cos(36.9)
And I think the force's on B would be [tex]\sum[/tex]Fy = 50 - T = massB*accelB
Now, that leaves the lengths of cord, which should give two equations as it's two separate cords. I believe the length of the cord directly attached to A will be Sa + 2Sc = L1
But I'm unsure of the second cord so here's where I'm stuck. Also I'm not entirely sure everything else I've done is right either so if someone wouldn't mind checking that would be great.