# Normal force acting on a block on an accelerating wedge

• Leo Liu
In summary: Those are in the wedge's frame of reference. But yeah, I'm not sure if that relative acceleration formula in my first post is usable in this problem.Please ignore my posts; the wedge is not an inertial frame of reference.
Leo Liu
Homework Statement
I created this stupid question to help me better understand Newtonian mechanics.
Relevant Equations
N/A

We have a wedge whose surface is ##\theta## from the horizontal surface. After a block is placed on its frictionless slant surface, the wedge starts to accelerate due to a force F. What is the normal force acting upon the block?

I have been trying to solve it but I got no clue. Could someone give me a hint?

First step: draw separate FBDs for the two components, showing forces. Next, create variables for the accelerations. Third, write the ΣF=ma equations and any appropriate kinematic equations that express relationships between the accelerations.
For that last, you need to represent the fact that the block stays on the wedge surface.

Leo Liu
haruspex said:
First step: draw separate FBDs for the two components, showing forces. Next, create variables for the accelerations. Third, write the ΣF=ma equations and any appropriate kinematic equations that express relationships between the accelerations.
For that last, you need to represent the fact that the block stays on the wedge surface.

Hi. The constraints in this problem gives these four distinct equations. Could you tell me if they are correct?

$$\begin{pmatrix} a_{sx}\\a_{sy}\\a_t\\N \end{pmatrix}=\begin{pmatrix} 0\\0\\-mg\\F \end{pmatrix}\begin{pmatrix} \tan \theta & 1 & 0 & 0\\ m & 0 & m & -sin \theta\\ 0 & m & 0 & -\cos \theta\\ 0 & 0 & M & \sin \theta \end{pmatrix}^{-1}$$

Last edited:
Is asx the horizontal acceleration of the block in the lab frame or in the frame of the wedge?

If you define the height and leg of the wedge as ##h## and ##\ell##, then the net force on the block in the wedge's reference frame is in the direction of ##(\ell, -h)##. Since we know that the net force's norm is ##g\sin\theta##, we can describe it by ##\vec f_{b|w} = \frac{g\sin\theta}{\sqrt{\ell^2+h^2}}(\ell, -h)##. I think that this might work (you can find the block's absolute acceleration from this since you know the wedge's acceleration).$$\vec a_{b|w}=\vec a_b-\vec a_w$$The block starts falling from the top, so i think that it okay that we are considering the whole wedge to be only the topmost point (so that ##\vec x_b(0)-\vec x_w(0)=\vec y_b(0)-\vec y_w(0)=\vec 0##).

Last edited:
haruspex said:
Is asx the horizontal acceleration of the block in the lab frame or in the frame of the wedge?
The frame of the wedge; since it is an non inertial frame, I added the acceleration of the frame to get the real acceleration.

Leo Liu said:
The frame of the wedge; since it is an non inertial frame, I added the acceleration of the frame to get the real acceleration.
Then it all looks fine.

Leo Liu
archaic said:
the net force's norm is ##g\sin\theta##
Can you please tell me why this isn't ##\mu N##? I don't think you can use the component of the weight vector to calculate the sliding friction force.

Leo Liu said:
Can you please tell me why this isn't ##\mu N##? I don't think you can use the component of the weight vector to calculate the sliding friction force.
Oh, that isn't the frictional force. You mentioned that your surface is frictionless, right? I meant by ##\vec f## the net force on the block.

The normal force would be in the direction of ##(\sin\theta,\cos\theta)## and with norm ##mg\cos\theta##, so:$$\vec N=mg\cos\theta(\sin\theta,\cos\theta)$$The force I have written in my other post has a missing ##m## factor. It should be:$$\vec f_{b|w}=mg\sin\theta(\cos\theta,-\sin\theta)$$

archaic said:
The normal force would be in the direction of ##(\sin\theta,\cos\theta)## and with norm ##mg\cos\theta##, so:$$\vec N=mg\cos\theta(\sin\theta,\cos\theta)$$The force I have written in my other post has a missing ##m## factor. It should be:$$\vec f_{b|w}=mg\sin\theta(\cos\theta,-\sin\theta)$$
I don't think this is the case since the wedge is accelerating.

Leo Liu said:
I don't think this is the case since the wedge is accelerating.
Those are in the wedge's frame of reference. But yeah, I'm not sure if that relative acceleration formula in my first post is usable in this problem.

Please ignore my posts; the wedge is not an inertial frame of reference.

## 1. What is the definition of normal force?

The normal force is the force exerted by a surface on an object that is in contact with it. It is always perpendicular to the surface and prevents the object from passing through the surface.

## 2. How is the normal force related to the weight of the object?

The normal force is equal in magnitude and opposite in direction to the weight of the object. This means that if the weight of the object increases, the normal force will also increase.

## 3. How does the normal force change when the block is on an accelerating wedge?

The normal force on the block will change depending on the direction and magnitude of the acceleration of the wedge. If the wedge is accelerating upwards, the normal force will decrease, and if the wedge is accelerating downwards, the normal force will increase.

## 4. What is the relationship between the normal force and the angle of the wedge?

The normal force is directly proportional to the cosine of the angle of the wedge. This means that as the angle of the wedge increases, the normal force will decrease.

## 5. How does the normal force affect the motion of the block on the wedge?

The normal force does not directly affect the motion of the block on the wedge. It is only responsible for preventing the block from falling through the wedge. The motion of the block is determined by the net force acting on it, which includes the normal force, weight, and any other forces present.

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