Two boxes sliding down inclined plane

• zeralda21
In summary: F_{T}=maF_{T}=m+gFor block 1:F_{net}=ma=-mgsin(30)+F_{T} which yields a=-\frac{g}{2}+\frac{F_{T}}{m}.For block 2:F_{net}=ma=-mgsin(30)+F_{T} which yields a=-\frac{g}{2}+\frac{F_{T}}{M}.So for block 1:F_{net}=ma=-mgsin(30)+F_{T} which yields a=-\frac{g}{2}+\frac{36
zeralda21

Homework Statement

A block with m=20kg is sliding on top of a block of M=10 kg that is sliding down an inclined plane with 30 degrees to the horizontal. All surfaces are frictionless. Calculate the acceleration of the blocks.

Homework Equations

Newtons second law: $$F⃗ net=ΣF⃗ =ma⃗$$

The Attempt at a Solution

For sake of generality, let's say that the angle is indefinite.
I have sketched a free-body diagram and $$F⃗ net=-mgsin(\theta)-Mgsin(\theta)=(m+M)a⃗$$ since the boxes will moves as one unit. Solving for acceleration gives
$$a⃗=-gsin(\theta)$$ which is wrong. I think it is something with the signs. They both have the same sign since they move in the same direction(downward=negative) but even if both positive would yield wrong answer.

Why do you think your answer is wrong? (They probably just want the magnitude of the acceleration, not the direction.)

it is correct. But think about what you have calculated. The total force on both the blocks, so this will give you the acceleration of what? (Hint: it is not the acceleration of one or other of the blocks).

Yes, magnitude is of importance here. In this case, with an angle 30 degrees, the acceleration would be $$\frac{g}{2}$$ which still is a wrong solution. The equation is supposed to be

$$(m+M)a=(m-M)gsin(\theta)$$ which in the end yields $$\frac{1}{6}g$$ but I fail to see why the forces are in the opposite direction of each other.

I guess I don't understand the problem. Is the incline fixed in place?

If this is from a textbook, can you give the name and problem number.

haha, the pulley makes a big difference :) Also, I think your answer before would have been right (as Doc Al said), but the inclusion of the pulley makes it a bit more complicated.

Now we have a completely different problem (from the one I was envisioning).

Start by drawing free body diagrams for each block.

Yes, that was a mistake of mine. Alright, the free body diagrams. Both are being pulled with a Force from the pulley, tension, call it $$F_{T}$$ There is also a gravitational force in the -y direction with magnitude mg. This can(for both blocks) be written $$mgsin(\theta)$$ which is parallel to the incline.
Also, for both blocks, there is a normal force, perpendicular to the incline.

So both free-body diagrams include the same forces, but with different magnitudes.

zeralda21 said:
Yes, that was a mistake of mine. Alright, the free body diagrams. Both are being pulled with a Force from the pulley, tension, call it $$F_{T}$$
Good.
There is also a gravitational force in the -y direction with magnitude mg. This can(for both blocks) be written $$mgsin(\theta)$$ which is parallel to the incline.
Good. Of course the blocks have different masses. (Personally, I would choose the positive direction to be down the incline, but it doesn't matter as long as you are consistent.)
So both free-body diagrams include the same forces, but with different magnitudes.
OK. Now apply Newton's 2nd law to each. (Careful with signs.)

Alright. I actually not sure how to include the normal force. Is it $$F_{N}sin(120)$$? For block 1: $$F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)$$
For block 2: $$F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)$$

zeralda21 said:
Alright. I actually not sure how to include the normal force. Is it $$F_{N}sin(120)$$? For block 1: $$F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)$$
For block 2: $$F_{net}=ma=-mgsin(30)+F_{T}+F_{N}sin(120)$$
The normal force acts perpendicularly to the direction of motion. Since there's no friction, you don't need to worry about it. And you would never add perpendicular force components!

Be sure to use different symbols for the masses. What's the relationship between the acceleration of each block?

Yes of course, i should have known that. If it is perpendicular, angle is 90 degrees and sin(90)=0. We label the mass of block 1, m, and mass of block 2 by M.

So for block 1:

$$F_{net}=ma=-mgsin(30)+F_{T}$$ which yields $$a=-\frac{g}{2}+\frac{F_{T}}{m}$$.

For block 2:

$$F_{net}=Ma=-Mgsin(30)+F_{T}$$ which yields $$a=-\frac{g}{2}+\frac{F_{T}}{M}$$.

So I will have to find $$F_{T}?$$ Is that possible even if the system is not in equilibrium?

Careful with the sign of the acceleration.

Let "a" stand for the magnitude of the acceleration. Pick one of the boxes (say M) and assume it slides down the incline. What would be its acceleration? What would be the acceleration of the other box?

As far as the tension is concerned, you have two equations and two unknowns (tension and acceleration). So there shouldn't be a problem in solving for both.

I fail too see why they should'nt have the same sign. If the lower block is moving with an acceleration a, and the other block moves exactly the same on top of it. Is not the acceleration the same for this block too? At least that is what my intuition tells me. Maybe I could think that one of them is 0 but different sign? I can't see that actually..

They cannot both move downwards, the string across the pulley has a fixed length. If one block goes down, the other one has to go up - and the question is, how are the two accelerations related (via the fixed string length).

mfb said:
They cannot both move downwards, the string across the pulley has a fixed length. If one block goes down, the other one has to go up - and the question is, how are the two accelerations related (via the fixed string length).

Of course. Oh well, that explains it. Thanks Doc Al and mrb.

1. How does the angle of the inclined plane affect the speed of the boxes?

The angle of the inclined plane affects the speed of the boxes because it determines the force of gravity acting on the boxes. The steeper the angle, the greater the force of gravity, and therefore the faster the boxes will accelerate down the plane.

2. How does the mass of the boxes affect their motion down the inclined plane?

The mass of the boxes affects their motion down the inclined plane because it determines the amount of inertia they have. Heavier boxes will have more inertia and will require more force to accelerate down the plane compared to lighter boxes.

3. What is the relationship between the friction coefficient and the acceleration of the boxes down the inclined plane?

The friction coefficient is a measure of the resistance to motion between two surfaces. The higher the coefficient, the more friction there is between the boxes and the inclined plane, which will slow down the acceleration of the boxes. Conversely, a lower friction coefficient will result in a faster acceleration down the plane.

4. How does the length of the inclined plane affect the motion of the boxes?

The length of the inclined plane does not affect the motion of the boxes as long as the angle of the plane remains constant. The force of gravity and the mass of the boxes will determine their acceleration, regardless of the length of the plane.

5. How can the final velocity of the boxes be calculated after reaching the bottom of the inclined plane?

The final velocity of the boxes can be calculated using the equation v = √(2gh), where v is the final velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the inclined plane. This equation assumes no friction and that the boxes start from rest at the top of the inclined plane.

Replies
11
Views
3K
Replies
4
Views
466
Replies
18
Views
3K
Replies
12
Views
1K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
8
Views
2K
Replies
16
Views
3K
Replies
13
Views
5K
Replies
16
Views
4K