Calculating Normal Force on a Wedge with Friction: Basic Dynamics"

• Clara Chung
In summary, for a block of mass 5 kg resting on a wedge placed on a table, with the wedge having a mass of 10 kg and an incline of 30 degrees, the normal reaction of the table acting on the wedge is 150N. For part (b), where the block slides down the wedge without friction and the wedge remains stationary, the normal reaction is 138N. For part (c), where there is a friction of 10 N between the block and the wedge surfaces and the wedge remains stationary, the normal reaction is 143N. To solve for (b) and (c), one can consider the forces acting on the block and wedge separately, and use the fact that the net force on the
Clara Chung

Homework Statement

A block of mass 5 kg rests on a wedge placed on a table. The mass of the wedge is 10 kg. The wedge has an incline of angle θ = 30 degrees.

a) Find the normal reaction of the table acting on the wedge.
ANS : 150N

b) Repeat (a) if the block slides down the wedge without friction. The wedge remains stationary all the time.
ANS : 138N

c)Repeat (a) if there is a friction of 10 N between the block and the wedge surfaces. The wedge still remains stationary.
ANS : 143N

F= ma

The Attempt at a Solution

I have no problem with (a), it is simply (5+10)x10 = 150N
But starting from (b), I have no ideas. I don't know how to obtain the net force when it is inclined by 30 degrees. I tried to separate 50N into 2 components, the component parallel to the surface of the wedge is canceled out. Only the component perpendicular to the surface remains. Then I separate it to 2 components again, and take the component perpendicular to x-axis into account. So it is 50 x cos 30 x cos 30 = 37.5 N, so b= 37.5 +100 = 137.5N. I don't know if my concept is correct or not.
I hope you can teach me b and c part

For (b) and (c), one way to do it is to consider the forces acting on the block alone, and on the wedge alone. That is, draw a free-body diagram of just the block, showing the forces acting on it, and another diagram showing just the wedge and the forces acting on it.

The wedge doesn't move at all, so the net force on the wedge must always be zero. Can you use that information to relate the normal force acting on the wedge from the table to any other forces that act on the wedge?

As for the block, in part (b) it slides down the wedge without friction. So, what is the direction of the net force? And what do you know about the accelerations of the block in directions parallel and perpendicular to the surface of the wedge? Can you use this information to work out how the block and the wedge surface are interacting? Last step: how does that interaction affect the normal force of the table acting on the wedge?

Clara Chung
James R said:
For (b) and (c), one way to do it is to consider the forces acting on the block alone, and on the wedge alone. That is, draw a free-body diagram of just the block, showing the forces acting on it, and another diagram showing just the wedge and the forces acting on it.

The wedge doesn't move at all, so the net force on the wedge must always be zero. Can you use that information to relate the normal force acting on the wedge from the table to any other forces that act on the wedge?

As for the block, in part (b) it slides down the wedge without friction. So, what is the direction of the net force? And what do you know about the accelerations of the block in directions parallel and perpendicular to the surface of the wedge? Can you use this information to work out how the block and the wedge surface are interacting? Last step: how does that interaction affect the normal force of the table acting on the wedge?

For the wedge, the force acting on it are Mg and a normal force perpendicular to the surface, also friction. So Nsinθ=f and Ncosθ + Mg = normal force from the table acting on the wedge...(1).
For the block, the force are mg and the normal force acting on the block by the wedge. Because there is only net force parallel to the surface of the wedge, so mg cosθ =N...(2). So, by combining them, mg cosθ cosθ + Mg = answer, am I right?

Clara Chung said:
For the wedge, the force acting on it are Mg and a normal force perpendicular to the surface, also friction. So Nsinθ=f and Ncosθ + Mg = normal force from the table acting on the wedge...(1).
For the block, the force are mg and the normal force acting on the block by the wedge. Because there is only net force parallel to the surface of the wedge, so mg cosθ =N...(2). So, by combining them, mg cosθ cosθ + Mg = answer, am I right?

and for c part, using your method, there is a new friction acting on the wedge, so for the wedge part, the equation would be Ncosθ + mg +10 sin 30 = 142.5 degrees?

Clara Chung said:
For the wedge, the force acting on it are Mg and a normal force perpendicular to the surface, also friction. So Nsinθ=f and Ncosθ + Mg = normal force from the table acting on the wedge...(1).
For the block, the force are mg and the normal force acting on the block by the wedge. Because there is only net force parallel to the surface of the wedge, so mg cosθ =N...(2). So, by combining them, mg cosθ cosθ + Mg = answer, am I right?
That looks right to me.

Take the mass of the block to be ##m## and the mass of the wedge to be ##M##.

In part (a), if the normal force on the block from the wedge is ##N##, then the vertical force on the block is ##N \cos \theta - mg =0##. In part (a), the wedge doesn't move, so this net vertical force is zero. This is not true in (b) and (c).

As for the wedge, the net vertical force on it is ##-N\cos \theta - Mg + N_t=0##, where ##N_t## is the force from the table. Since the wedge doesn't accelerate, the net force is zero, so ##N_t = N\cos \theta + Mg##. This is true for all parts of the problem, since the wedge never accelerates.

For part (a), we can combine results to get ##N_t = mg + Mg = (m+M)g##, which is what you calculated originally, and which makes sense because the table must support both the wedge and the stationary block.

For (b) and (c) it is no longer true that the vertical force on the block is zero, but we do know that in both cases the acceleration of the block perpendicular to the surface of the wedge is zero. That is, ##N - mg\cos\theta = 0##, or ##N=mg\cos\theta##. The wedge still doesn't accelerate, so the expression for the normal force of the table in terms of ##N## is the same as before (but the ##N## has changed). Therefore we get ##N_t = N\cos\theta + Mg = (mg\cos\theta)\cos\theta + Mg = mg\cos^2 \theta + Mg##, which was your result, too. I don't think it makes any difference whether there is friction between the wedge and the block as it slides, because the friction force has no effect on the normal force between the wedge and the block. Do you agree?

Clara Chung
James R said:
That looks right to me.

Take the mass of the block to be ##m## and the mass of the wedge to be ##M##.

In part (a), if the normal force on the block from the wedge is ##N##, then the vertical force on the block is ##N \cos \theta - mg =0##. In part (a), the wedge doesn't move, so this net vertical force is zero. This is not true in (b) and (c).

As for the wedge, the net vertical force on it is ##-N\cos \theta - Mg + N_t=0##, where ##N_t## is the force from the table. Since the wedge doesn't accelerate, the net force is zero, so ##N_t = N\cos \theta + Mg##. This is true for all parts of the problem, since the wedge never accelerates.

For part (a), we can combine results to get ##N_t = mg + Mg = (m+M)g##, which is what you calculated originally, and which makes sense because the table must support both the wedge and the stationary block.

For (b) and (c) it is no longer true that the vertical force on the block is zero, but we do know that in both cases the acceleration of the block perpendicular to the surface of the wedge is zero. That is, ##N - mg\cos\theta = 0##, or ##N=mg\cos\theta##. The wedge still doesn't accelerate, so the expression for the normal force of the table in terms of ##N## is the same as before (but the ##N## has changed). Therefore we get ##N_t = N\cos\theta + Mg = (mg\cos\theta)\cos\theta + Mg = mg\cos^2 \theta + Mg##, which was your result, too. I don't think it makes any difference whether there is friction between the wedge and the block as it slides, because the friction force has no effect on the normal force between the wedge and the block. Do you agree?
James R said:
That looks right to me.

Take the mass of the block to be ##m## and the mass of the wedge to be ##M##.

In part (a), if the normal force on the block from the wedge is ##N##, then the vertical force on the block is ##N \cos \theta - mg =0##. In part (a), the wedge doesn't move, so this net vertical force is zero. This is not true in (b) and (c).

As for the wedge, the net vertical force on it is ##-N\cos \theta - Mg + N_t=0##, where ##N_t## is the force from the table. Since the wedge doesn't accelerate, the net force is zero, so ##N_t = N\cos \theta + Mg##. This is true for all parts of the problem, since the wedge never accelerates.

For part (a), we can combine results to get ##N_t = mg + Mg = (m+M)g##, which is what you calculated originally, and which makes sense because the table must support both the wedge and the stationary block.

For (b) and (c) it is no longer true that the vertical force on the block is zero, but we do know that in both cases the acceleration of the block perpendicular to the surface of the wedge is zero. That is, ##N - mg\cos\theta = 0##, or ##N=mg\cos\theta##. The wedge still doesn't accelerate, so the expression for the normal force of the table in terms of ##N## is the same as before (but the ##N## has changed). Therefore we get ##N_t = N\cos\theta + Mg = (mg\cos\theta)\cos\theta + Mg = mg\cos^2 \theta + Mg##, which was your result, too. I don't think it makes any difference whether there is friction between the wedge and the block as it slides, because the friction force has no effect on the normal force between the wedge and the block. Do you agree?

Thanks for your answer for (a) part, I finally understand why the force components are always set parallel/perpendicular to the motion.
But for c part, I think the friction force has indeed no effect on the normal force on the block from the wedge, because they are perpendicular, but when I deal with the wedge, the friction is not perpendicular to the normal force given to the wedge by the table, so friction should be taken to account along with the normal force from the block, mass of the wedge?

James R said:
As for the block, in part (b) it slides down the wedge without friction. So, what is the direction of the net force? And what do you know about the accelerations of the block in directions parallel and perpendicular to the surface of the wedge?
Clara did all that, correctly, in the original post.
James R said:
I don't think it makes any difference whether there is friction between the wedge and the block as it slide
It does. Sufficient friction would have the block stationary, so would be as in part a.
Clara Chung said:
friction should be taken to account along with the normal force from the block, mass of the wedge?
Yes. what do you think its contribution will be?

Clara Chung
James R said:
I don't think it makes any difference whether there is friction between the wedge and the block as it slide
haruspex said:
It does. Sufficient friction would have the block stationary, so would be as in part a.
Yikes! You're right, of course. Thanks haruspex.

Sorry Clara. This is what happens when you try to do this sort of problem completely in your head. The best first step in approaching these kinds of problems is to draw some free-body diagrams showing all the forces acting, with the relevant angles and directions. That's what I should have done first up, but lazily didn't.

In this case, it is helpful to draw separate free-body diagrams for the block and for the wedge. Keep Newton's third law in mind to get the correct directions for the forces exerted by the block on the wedge.

So, your solutions (and mine) to parts (a) and (b) above are correct, but I was wrong about some of the things I said about part (c). In part (c), you should have a total of 4 separate forces acting on the wedge, and you're asked to find the magnitude of one of them, namely the normal force from the table. We are told the wedge remains stationary, so the net force on the wedge must be zero. It also follows that the sum of the components of the 4 forces in any chosen direction must be zero, too.

So, can you write an expression for the net force in terms of all of the forces acting on the wedge, which will allow you to calculate the normal force on the wedge from the table?

1. How do I calculate the normal force on a wedge with friction?

To calculate the normal force on a wedge with friction, you will need to use the formula: Normal Force = weight of the object x cos (angle of the wedge). This formula takes into account the weight of the object and the angle of the wedge, which affects the normal force due to the force of gravity.

2. What is the role of friction in calculating normal force on a wedge?

Friction plays a crucial role in calculating normal force on a wedge. Friction is the force that opposes the motion of an object, and it is present whenever two surfaces are in contact. In the case of a wedge, friction helps to keep the object from sliding down the wedge, which affects the normal force that is exerted on the object.

3. How does the angle of the wedge affect the normal force?

The angle of the wedge has a direct impact on the normal force. The normal force is equal to the weight of the object multiplied by the cosine of the angle of the wedge. This means that as the angle of the wedge increases, the normal force decreases, and vice versa.

4. What is the significance of calculating the normal force on a wedge with friction?

Calculating the normal force on a wedge with friction is important because it helps to determine the stability of the object on the wedge. The normal force is the force that is perpendicular to the surface of the wedge, and it is responsible for keeping the object in place and preventing it from sliding down the wedge.

5. What are the units for normal force?

The units for normal force are newtons (N) in the International System of Units (SI). However, in the British Gravitational System (BG), the unit for normal force is pounds (lb).

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