What is the magnitude of acceleration?

[Adriano25]

Homework Statement

M1 = 7.0 kg
M2 = 3.0 kg
Coefficient of friction between two blocks: 0.4
Angle of incline: 35 degrees
Ramp is friction less

Homework Equations

For Mass 1:
ΣF1x = m1a1
-m1gsinθ + T + μkN = m1a1 (1)
ΣF1y = 0
N = m1gcosθ (2)

For Mass 2:
ΣF2x = m2a2
-m2gsinθ + T - μkN = m2a2 (3)
ΣF1y = 0
-m2gcosθ + N + N1 = 0 (4)

The Attempt at a Solution

I assumed both accelerations are the same. Thus a1=a2 ⇒ a

I subtracted equations (1) & (3)
-m1gsinθ + m2gsinθT + 2μkN = m1a - m2a (5)
(2) into (5) and solve for acceleration which gives me 2.25 m/s²

Does it look right? Also, how can I know if the speed of m1 is increasing or decreasing?

 

[Simon Bridge]

I assumed both accelerations are the same. Thus a1=a2 ⇒ a

Their magnitudes are the same - acceleration is a vector. In your analysis, the direction of the vector is indicated by the sign. You appear to define "up the slope" as positive for both free body diagrams (clue: T>0) ... in that case, a1 = -a2... because the blocks will accelerate in opposite directions. How did you account for this?
You also appear to have assumed that m1 is going down the slope (off the direction of the friction force) - why?

Also, how can I know if the speed of m1 is increasing or decreasing?

If the acceleration is opposite to the velocity, then the speed is decreasing.

 

[TSny]

Are the blocks released from rest?

 

[Adriano25]

Their magnitudes are the same - acceleration is a vector. In your analysis, the direction of the vector is indicated by the sign. You appear to define "up the slope" as positive for both free body diagrams (clue: T>0) ... in that case, a1 = -a2... because the blocks will accelerate in opposite directions. How did you account for this?
You also appear to have assumed that m1 is going down the slope (off the direction of the friction force) - why?
If the acceleration is opposite to the velocity, then the speed is decreasing.

Hi Simon,
Thank you for taking the time to help on this. I forgot to mention that the problem states that m1 is moving down the incline. Therefore, if I define up the slope as positive, it means that acceleration of m1 will be negative correct? so the relationship between accelerations of m1 and m2 will be, -a1=a2?

 

[Adriano25]

Are the blocks released from rest?

Hi TSny,
The problem does not mention this information. It states that two blocks move on a 35 degree incline. Also, it states that m1 is moving down the incline.

 

[Simon Bridge]

Hi Simon,
Thank you for taking the time to help on this. I forgot to mention that the problem states that m1 is moving down the incline. Therefore, if I define up the slope as positive, it means that acceleration of m1 will be negative correct?

Only if m1 is speeding up.

...so the relationship between accelerations of m1 and m2 will be, -a1=a2?

It does not matter where the minus sign goes in that relation as, here, it indicates the relative direction of the accelerations of the blocks, not the acceleration wrt the coordinates. -a1=a2 means that when a1 points down, a1 points up ... a1=-a2 says the same thing.

The problem does not mention this information [if the blocks were "released from rest"]. It states that two blocks move on a 35 degree incline. Also, it states that m1 is moving down the incline.

ie. the problem statement says that the blocks are not at rest at the time shown, thus: $|v_1| > 0$, and $\vec v_1$ points down the incline. (The time shown may not have been the time released, but that does not matter.)

Thus: if the acceleration is positive, $m_1$ is slowing down.

What would the forces have to be like for m1 to slow down?

 

[Adriano25]

Only if m1 is speeding up. It does not matter where the minus sign goes in that relation as, here, it indicates the relative direction of the accelerations of the blocks, not the acceleration wrt the coordinates. -a1=a2 means that when a1 points down, a1 points up ... a1=-a2 says the same thing.
ie. the problem statement says that the blocks are not at rest at the time shown, thus: $|v_1| > 0$, and $\vec v_1$ points down the incline. (The time shown may not have been the time released, but that does not matter.)

Thus: if the acceleration is positive, $m_1$ is slowing down.

What would the forces have to be like for m1 to slow down?

They would have to be positive then since the velocity vector is pointing down (negative direction) and the acceleration found is pointing up (positive direction). Is that right? Also, I'm still not confident with my answer 2.25 m/s² since I keep getting two different tension forces using the equations for M1 and M2. Any thoughts? Thank you for your help.

 

[Simon Bridge]

What would the forces have to be like for m1 to slow down?

They would have to be positive then since the velocity vector is pointing down (negative direction) and the acceleration found is pointing up (positive direction).

... "they" being the gravitational and friction force?
If they are both positive, and positive is up, then they both point up. How can the gravity force point up?

 

[Adriano25]

... "they" being the gravitational and friction force?
If they are both positive, and positive is up, then they both point up. How can the gravity force point up?

I apologize I misunderstood your question. Gravity would be negative, friction positive, and tension positive.
ΣF1x = m1a1
-m1gsinθ + T + μkN = m1a1
where N = m1gcosθ

I think I figured it out. Thank you very much. I'm pretty confident that my answer is a=5.63 m/s². Therefore, I get a tension of 56.3 N. Since the result of acceleration is positive, m1's speed is decreasing because its velocity vector is negative, i.e., pointing in the opposite direction.

 

[haruspex]

my answer is a=5.63 m/s².

seems too much. Please post your working.

 

[Adriano25]

seems too much. Please post your working.

For Mass 1:
ΣF1x = m1a1
-m1gsinθ + T + μkN = m1a1 (1)
ΣF1y = 0
N = m1gcosθ (2)

For Mass 2:
ΣF2x = m2a2
-m2gsinθ + T - μkN = m2a2 (3)
ΣF1y = 0
-m2gcosθ + N + N1 = 0 (4)

Therefore,
I subtracted equations (1) & (3)
-m1gsinθ + m2gsinθ + 2μkN = m1a - m2a (5)
(2) into (5)
-m1gsinθ + m2gsinθ + 2μk(m1gcosθ) = m1a - m2a
a = [-m1gsinθ + m2gsinθ + 2μk(m1gcosθ)] / m1-m2
a = 5.63 m/s²

Does my work look right? Please any suggestions would be very helpful.
Thank you

 

[haruspex]

-m1gsinθ + m2gsinθ + 2μkN = m1a - m2a (5)

As has already been pointed out to you, the two accelerations are in opposite directions. Since you have made up the slope positive for a1 and a2, you have a1=-a2.

 

[Adriano25]

As has already been pointed out to you, the two accelerations are in opposite directions. Since you have made up the slope positive for a1 and a2, you have a1=-a2.

Thank you. I got a1 = 2.25 m/s². Since it's positive, m1's speed is decreasing.

 

[haruspex]

Thank you. I got a1 = 2.25 m/s². Since it's positive, m1's speed is decreasing.

Looks right.