Acceleration on incline (w. Friction)

[TomFoolery]

Homework Statement

- Taken directly from the homework -

- Attachment has the diagram -

Body A in Fig. 6-33 weighs 100 N, and body B weighs 78 N. The coefficients of friction between A and the incline are μs = 0.54 and μk = 0.22. Angle θ is 29°. Let the positive direction of an x-axis be down the slope. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

The Attempt at a Solution

My problem is in figuring out which equations to setup. My initial thoughts are as follow:

a) figure out the acceleration of the system, then factor in friction after finding which direction the block wants to travel.

F_netA = sin(29)100N - T
F_netB = T - 78N

b) This didnt account for friction right off, so I modified for this:

F_netA = sin(29)100N - T - (100N)(0.54)(cos(29))
F_netB = T - 78N

a & b) Once having these equations, I set them up to equal T, then set F_netA(solved for T) equal to F_netB(solved for T).

This still didn't really seem to yield any insight. Please help me realize how to setup these initial equations and I'm sure the rest will fall in place.

 

[PhanthomJay]

Homework Statement

- Taken directly from the homework -

- Attachment has the diagram -

Body A in Fig. 6-33 weighs 100 N, and body B weighs 78 N. The coefficients of friction between A and the incline are μs = 0.54 and μk = 0.22. Angle θ is 29°. Let the positive direction of an x-axis be down the slope. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

The Attempt at a Solution

My problem is in figuring out which equations to setup. My initial thoughts are as follow:

a) figure out the acceleration of the system, then factor in friction after finding which direction the block wants to travel.

No, you must factor in friction first, it is given which way the block is travelling, except when it is not travelling, in which case you have to figure out the friction force direction.

F_netA = sin(29)100N - T

no

F_netB = T - 78N

yes, if the block on the plane is moving down and the other is moving up, or if neither is moving

b) This didnt account for friction right off, so I modified for this:

F_netA = sin(29)100N - T - (100N)(0.54)(cos(29))

no, you apparently are looking at the stationary case, since you seem to be using the static friction coefficient. What is the acceleration of the blocks when they are at rest? What is the net force on each block when they are at rest? Bonus: What is the friction force acting on the block on the incline when it is at rest ? (Note:it is not uN!)

F_netB = T - 78N

yes...

a & b) Once having these equations, I set them up to equal T, then set F_netA(solved for T) equal to F_netB(solved for T).

This still didn't really seem to yield any insight. Please help me realize how to setup these initial equations and I'm sure the rest will fall in place.

there are 3 parts to this problem, look at that one at a time.

When the block is moving up the incline, what is the direction of the friction force, of the tension force, and of the weight force parallel to the incline?

When the block is moving down the incline, same questions.

When the blocks are not moving, same questions. In particular, what must be the direction of the friction force?

Then use Newton's laws on each block to solve for the acceleration.

 

[TomFoolery]

I thought that I had the idea, but only one part of it was correct. My approach is this:

Part A (initially stopped): since the block is stopped, then one of the forces (78N or sin(29)*100N) must overcome the other force AND the force of static friction. Neither one can do this, thus the acceleration is zero. [Got this part correct].

Part B (Moving UP the incline): F_netA = (100N)(sin(29)) + (0.22)(100N)(cos(29)) - 78N
In words, the force of gravity pulls 100N down the x-axis (positive direction) and the force of friction also faces down the x axis, while the force of tension pulls it up the x-axis (in the negative direction).
F_netA = 48.48N + 19.24N -78N = -10.28N
= -10.28N / (100N / 9.8m/s^2)
= -10.28N / 10.20kg = -1.0078 m/s^2
This is: the net force divided by mg/g (which gives mass of A), the end result should be acceleration.

Part C (Moving DOWN the incline): F_netA = (100N)(sin(29)) - (0.22)(100N)(cos(29)) - 78N
Same as above, but with the force of friction reversed, because the force is now facing up the incline.
F_netA = 48.48N - 19.24N - 78N = -48.76N
-48.76N / 10.20kg = -4.78 m/s^2

*I got part B and C wrong, I can't see why though.

 

[PhanthomJay]

I thought that I had the idea, but only one part of it was correct. My approach is this:

Part B (Moving UP the incline): F_netA = (100N)(sin(29)) + (0.22)(100N)(cos(29)) - 78N
In words, the force of gravity pulls 100N down the x-axis (positive direction) and the force of friction also faces down the x axis, while the force of tension pulls it up the x-axis (in the negative direction).

yes, very good

F_netA = 48.48N + 19.24N -78N = -10.28N
= -10.28N / (100N / 9.8m/s^2)
= -10.28N / 10.20kg = -1.0078 m/s^2
This is: the net force divided by mg/g (which gives mass of A), the end result should be acceleration.

You have an error here in your assumption that the tension force in the rope is 78 N. All you can say is that the tension force is T when you write your equation from Newton 2. You have 2 unknowns in that equation, T and a. You must now look at the hanging block and after identifying the forces acting on it, use Newton 2 for that block, getting an equation where T and a are the unknowns. Now solve the 2 simultaneous equations for T and a.

Part C (Moving DOWN the incline): F_netA = (100N)(sin(29)) - (0.22)(100N)(cos(29)) - 78N
Same as above, but with the force of friction reversed, because the force is now facing up the incline.

yes, very good

F_netA = 48.48N - 19.24N - 78N = -48.76N
-48.76N / 10.20kg = -4.78 m/s^2

*I got part B and C wrong, I can't see why though.

Same error in part C.

 

[TomFoolery]

I've got it now, thank you for all of your help. My problem was that I was assuming the Net force on B to still be zero. I should have resolved for T and still included mass of B times acceleration of the system... rather simple in hindsight.