# Block on a frictionless, accelerating, inclined plane

• chronos98
In summary, Jared calculates the acceleration of a 2 kg block resting on a 60 deg incline and an acceleration of 16.9914 m/s/s.
chronos98

## Homework Statement

A 2.0 kg block rests on a frictionless wedge that has a 60 deg incline and an acceleration $$\vec{a}$$ to the right such that the mass remains stationary relative to the wedge. a) Draw the free-body diagram of the block and use it to determine the magnitude of the acceleration.

## Homework Equations

F=ma
3. Picture on right. My attempted diagram on left
http://img707.imageshack.us/img707/2420/physicsm.jpg

## The Attempt at a Solution

So my main problem is determining what forces are acting on the block, and drawing the diagram. When I tried it, I thought that the normal force would be increased due to the acceleration of the wedge, but I am certain I have done this wrong. So, I drew a diagram of forces acting on the block which I said were the normal force, perpendicular to the wedge, and the force of gravity acting downwards. My coordinate system was setup so gravity pointed in the -y direction, and x was pointing straight to the right. I said that $$\sum(F_x)=F_n*\cos(30)=(2.0 kg)*\vec{a}$$ and $$\sum(F_y)=F_n*\sin(30)-F_g=0$$. I used the summation of forces in y to find the normal force, which I plugged into the first to find acceleration. I got a normal force of 39.24N and an acceleration of 16.9914 m/s/s.

Last edited by a moderator:
Uh, we're going to need your question and attempt at the very least.

Jared

Sorry, I hit enter too soon

So you need to calculated the force parallel to the ramp.

F = (m*g) / sin(theta) = (2*9.8)/sin60

Now you have the force acting down the ramp, you can use F=ma to get the acceleration.

Jared

So you're saying that since gravity is the only force pulling the block down, the acceleration component must be opposite that? I understand that. using that equation, you get an acceleration down the ramp of 8.495709211 m/s/s. Using trigonometry, a = ax/cos(60) where ax is 80495719211 m/s/s. that gives me 16.9914 m/s/s which is the same answer I already got. So is that the right answer?

F = (m*g) / sin(theta) = (2*9.8)/sin60 = 16.97N of force acting down the ramp - this is the force acting down parallel to the ramp.

F = ma = 16.97 = 2*a

Rearrange

16.97 / 2 = 8.49m/s^2

Acceleration = 8.49m/s^2

I don't know why you do that final trig at the end. using sin(theta) above compensates for the slope of the ramp.

Your free body diagram needs some modification to show the force acting parallel to the ramp downwards, which is what is / you are using to calculate acceleration.

A useful tool that will save you when you get into more complex physics is called "translating your coordinate system". You can put your coordinate system anyway you want, it does not change the physics of the problem. With that said, if you place your x-axis on the slope of the triangle block and your y-axis perpendicular to your x, having your block at the origin, than you can simplify the problem a good amount.

[PLAIN]http://img801.imageshack.us/img801/3856/blocke.png

Summing the forces is quite easier isn't it

Forces in Y = N - mgcos(t) = 0

Force in x = mgsin(t) = ma;

now solve

Last edited by a moderator:
Thanks fizzynoob, like I said, he should also show the force acting down the ramp as that is what you are using to calculate the acceleration. Just me being picky with the drawing styles.

>>force acting down the ramp
Did i not incorporate that with the force of gravity? gravity is the only force that has an x component.

>>Just me being picky with the drawing styles.
Yeah i realize that everyone on here has there own style, which corresponding to there professors style. It would be nice if all the professors would just create a standard so there's no confusion

There is a standard, the force acting down the ramp does involve gravity, but it is the amount of force out of the weight which acts parallel down the ramp.

This link shows the forces of such a block on a ramp:
http://myweb.cwpost.liu.edu/vdivener/notes/angle_of_repose.htm

This is the way you should represent the vector forces. He doesn't need the normal force so needn't show it, but he's using the shear force and so it does need to be on the diagram.

Last edited by a moderator:
>>There is a standard, the force acting down the ramp does involve gravity, but it is the amount of force out of the weight which acts parallel down the ramp.

yes which i put

>>Force in x = mgsin(t) = ma;

I didn't put the different components because it was just a FBD to show the magnitudes of the forces.

>>He doesn't need the normal force so needn't show it
It does not contribute because there no friction, but why start developing bad habits now? Its good practice to draw out the forces, even if they are irrelevant. Plus they help you understand the physics of the problem

Last edited:
I see your x now in the bottom. Apologies.

I understand that the 8.4957 m/s/s is the acceleration of the block down the ramp. I don't believe that is the acceleration that I want, though. The question is asking for the acceleration of the wedge parallel to the ground. That's why I did the trigonometry at the end

Last edited:
Well in that case, look at your FBD (free body diagram). The only force that contributes to the movement of the wedge in the x direction is from the normal of the block onto the wedge.

sum the forces in the x for the wedge

Forces in X = -Nsin(t) = -ma; // a being the acceleration of the wedge

Now the normal is relevant to the problem.

To find the normal you can sum the forces in the Y direction

Forces in the Y = Nsin(t) - mg = 0;

solve for N, than plug that normal into the force equation for the x direction

What about component of the acceleration parallel to the ramp? That will be the greatest here and should be less than the figure parallel to the ramp, not more. The acceleration figure we have already does not have the normal included but it has a horizontal component. Using my phone at the moment, will be back later hopefully.

I was referencing his fbd in his first post, translating the cooridnate system won't help when dealing with the wedge. Therefor the only component that results in the wedge moving is from the normal force due to the block

thats why I put, -Nsin(t) =-ma

Thanks for the replies. @Fizzynoob- that's the approach I took when I did it the first time, except I summed up forces for the block because the mass of the wedge is unknown.

Neither the block or the wedge is moving.

The question states an acceleration is in place to hold the block so it neither ascends or descends.

The acceleration we have calculated is how fast it would descend if allowed to freely drop down the ramp. To keep the block steady you would have to apply that acceleration in the opposite direction. In your drawing, the wedge is the wrong way around. It should be with the right angle corner on the right hand side, so the acceleration is pushing on the block up the ramp.

In this case the acceleration down the ramp is 8.4957m/s^2, so you would have to apply -8.4957m/s^2 up the ramp to compensate and keep it steady.

Now you need to calculate the right hand horizontal component. Remember, it has to compensate for the vertical acceleration also.

Last edited:
You can find the mass of tge wedge by summing the forces in the y direction I believe

The block or the wedge doesn't move. The acceleration applied has to counteract the acceleration down the ramp.

ah i see, i thought they wanted the acceleration as the block slides down, but they tell you that the block is stationary. So yes you are correct, the acceleration needed to keep the block stationary is equal to negative of the acceleration which it would be sliding down.

Now I'm going to make an assumption here: they tell you it is an acceleration to the right of the block, so I'd assume parallel to the ground. Which means you need a horizontal acceleration which generates the 8.xxxxm/s^2 acceleration at 60 degrees.

Sorry jarednjames, but that picture is correct. It's a copy out a picture given in the book. I went and talked to a physics professor today, and the way that I have it set up is correct. Sorry to waste all of your time.

In which case the acceleration is negative (if it's applied to the right).

When the force acts on the wedge along with the block, the acceleration of the system is
a = F/(M + m) in the horizontal direction towards right. Due to inertia the block accelerates in the opposite direction. Its component along the inclined plane is a*cosθ. This component prevents the downward sliding of the block with acceleration g*sinθ.
So the block will be at rest when a*cosθ = g*sinθ.

I think this is a poorly worded question. I don't see where it says the acceleration acts (at least not clearly).

A 2.0 kg block rests on a frictionless wedge that has a 60 deg incline and an acceleration LaTeX Code: \\vec{a} to the right

In the problem it is clearly stated that the wedge is accelerating to the right. So there must be a force acting on the wedge.

Well, I tried the problem my self, and as my instructor had mentioned in class, I took the direction of motion (i.e. direction of acceleration) to be the positive x-axis.

The weight of the 2.0 kg block is PERPENDICULAR to the direction of motion, but points downwards, so 'mg' is in the -ve y-direction.

If you follow the co-ordinate system mentioned here, you will have to resolve just the normal force Fn into its components.

Fn (along +ve-x axis) = Fn (cos 60)
Fn (along +ve-y axis) = Fn (sin 60)

Now, trying to sum up the forces along the +ve x-co-ordinate system,

{EQUATION 1} Total F(x) = Fn (cos 60) = Force applied F (ma)

Also, summing up the forces along the +ve y-co-ordinate system,

{EQUATION 2} Total F(y) = Fn (sin 60) - mg = 0

This F(y) is EQUAL to ZERO, since there is no acceleration in the y-direction.

On dividing equation 1 by equation 2, you get:

1/ [tan (60)] = (a/g)

The final answer comes out to be a = 5.66 m/s^2

I think it makes more sense if you draw out the FBD ... I could be wrong, but that is the only way it made some sense to me. :)

## 1. What is a block on a frictionless, accelerating, inclined plane?

A block on a frictionless, accelerating, inclined plane is a common physics problem that involves a block placed on an inclined plane with no friction, and the plane is accelerating. This problem helps to demonstrate the principles of Newton's second law and the equations of motion.

## 2. What is the acceleration of a block on a frictionless, accelerating, inclined plane?

The acceleration of a block on a frictionless, accelerating, inclined plane is equal to the acceleration due to gravity multiplied by the sine of the angle of inclination. This is known as the component of gravity parallel to the incline and it is the only force acting on the block in this scenario.

## 3. How does the angle of inclination affect the motion of a block on a frictionless, accelerating, inclined plane?

The angle of inclination has a direct effect on the acceleration of the block. As the angle increases, the acceleration of the block decreases. This is because the component of gravity parallel to the incline decreases as the angle increases, resulting in a smaller force acting on the block.

## 4. What is the significance of a frictionless surface in this problem?

The frictionless surface allows us to simplify the problem and focus on the effects of gravity on the block. In real-life scenarios, friction would play a significant role in the motion of the block, but by assuming a frictionless surface, we can isolate the effects of gravity and better understand the principles at play.

## 5. How can we calculate the velocity and displacement of a block on a frictionless, accelerating, inclined plane?

To calculate the velocity and displacement of a block on a frictionless, accelerating, inclined plane, we can use the equations of motion (v = u + at and s = ut + 1/2at^2) along with the acceleration due to gravity and the angle of inclination. By plugging in the known values and solving for the unknowns, we can determine the velocity and displacement of the block at any given time.

• Introductory Physics Homework Help
Replies
41
Views
460
• Introductory Physics Homework Help
Replies
12
Views
1K
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
836
• Introductory Physics Homework Help
Replies
30
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
948
• Introductory Physics Homework Help
Replies
2
Views
2K