M1/Physics Help: Have I a separated this correctly?

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TheAdviser101
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Am I right to assume this?
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TheAdviser101
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Lauren-x-
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I haven't seen it done like this before, though I would've separated mg into mgsina (parallel to the slope) and mgcosa (perpendicular to the slope) where a is theta. Did the question say you had to consider the box and the slope independently?

Which exam board is this for?
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RogerOxon
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Typically, you'd be wanting forces parallel (including friction, to calculate acceleration) and normal to the slope (for equilibrium in that direction), not vertical and horizontal.

What are you trying to calculate?
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atsruser
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(Original post by TheAdviser101)
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The components of R seem to be correct and mg is the correct weight of the block.

However, there is no mg force acting on the wedge. You need to think in terms of N3 force pairs. The mg force on the block is caused by the gravity of the Earth, so the corresponding paired force is the gravitational force of the block on the Earth - it is also of magnitude mg (by N3) and it pulls the Earth toward the block. You don't need to draw this, as you don't care about the motion of the Earth.

The N3 force pair between the block and the wedge is the reaction R of the wedge on the block, and the equal and opposite reaction force, of magnitude R, with which the block pushes on the wedge. However, the wedge doesn't feel the mg force at all - it can't since that is the force of gravity on a different object. It does however feel its *own* weight, Mg or whatever, but you haven't drawn this.

As others have mentioned, these aren't necessarily the most useful directions in which to resolve anyway since usually you want to find the motion of the block parallel to the slope. (They're fine if you want to find the motion of the wedge, however, but in M1 that's unlikely)
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Sir Cumference
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(Original post by TheAdviser101)
Am I right to assume this?
Just adding one thing to atsruser's excellent explanation:

For a block in equilibrium on a horizontal ground, the forces exerted on the block are its weight and the reaction of the ground on the block. To maintain equilibrium, the reaction of the ground on the block must have the same magnitude as the weight of the block.

But it's a common misconception to think that the block's weight and the reaction on the block are a N3 pair - they are not and it's just coincidental in a way that they have the same magnitude. Plus depending on the situation, the ground may not exert a force on the block that is equal in magnitude to the block's weight. This is the case in an accelerating lift for example.

This is why lift questions cause so many problems for students - the basics aren't taught properly by teachers.
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TheAdviser101
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(Original post by atsruser)
The components of R seem to be correct and mg is the correct weight of the block.

However, there is no mg force acting on the wedge. You need to think in terms of N3 force pairs. The mg force on the block is caused by the gravity of the Earth, so the corresponding paired force is the gravitational force of the block on the Earth - it is also of magnitude mg (by N3) and it pulls the Earth toward the block. You don't need to draw this, as you don't care about the motion of the Earth.

The N3 force pair between the block and the wedge is the reaction R of the wedge on the block, and the equal and opposite reaction force, of magnitude R, with which the block pushes on the wedge. However, the wedge doesn't feel the mg force at all - it can't since that is the force of gravity on a different object. It does however feel its *own* weight, Mg or whatever, but you haven't drawn this.

As others have mentioned, these aren't necessarily the most useful directions in which to resolve anyway since usually you want to find the motion of the block parallel to the slope. (They're fine if you want to find the motion of the wedge, however, but in M1 that's unlikely)
Thanks for your concise explanation. Btw this is for M1. My teacher gave me some extensions questions on mechanics and one of the questions involved a horizontal force on the block.

One thing that troubled me about this question was that if you split the reaction force into its real vertical component (r cos theta) and equate it to mg, the answer for the reaction force is wrong. Can you think of a possible reason why it does not work?
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TheAdviser101
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(Original post by Lauren-x-)
I haven't seen it done like this before, though I would've separated mg into mgsina (parallel to the slope) and mgcosa (perpendicular to the slope) where a is theta. Did the question say you had to consider the box and the slope independently?

Which exam board is this for?
It's part of a extension question for m1/physics. In this particular question there was a horizontal force acting on F. The force F acting on the wedge was just enough for the block to stay stationary relative to the wedge, which is in motion.
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atsruser
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(Original post by TheAdviser101)
One thing that troubled me about this question was that if you split the reaction force into its real vertical component (r cos theta) and equate it to mg, the answer for the reaction force is wrong. Can you think of a possible reason why it does not work?
There are only two forces acting on the block, mg and R. By N2, they determine its motion.

R is perp. to the slope, and thus has no component || to the slope. mg has component mg \sin\theta || to the slope. This is the only force acting || to the slope, so by N2, the block accelerates || to the slope (N2 says that a) an object accelerates if a nett force acts on it and b) the acceleration is in the direction of the nett force). Call this acceleration a_{||}

But a_{||} has a non-zero vertical component; call it a_v. Hence by N2, there must be a non-zero vertical force acting on the block. The vector sum of the vertical components of R and mg must thus cause this acceln, since they are the only forces with components in that direction. Hence we have:

mg-R\cos\theta = m a_v \ne 0 since a_v \ne 0

In short, you were assuming that the block was in vertical equilibrium, but it's not.
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TheAdviser101
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(Original post by atsruser)
There are only two forces acting on the block, mg and R. By N2, they determine its motion.

R is perp. to the slope, and thus has no component || to the slope. mg has component mg \sin\theta || to the slope. This is the only force acting || to the slope, so by N2, the block accelerates || to the slope (N2 says that a) an object accelerates if a nett force acts on it and b) the acceleration is in the direction of the nett force). Call this acceleration a_{||}

But a_{||} has a non-zero vertical component; call it a_v. Hence by N2, there must be a non-zero vertical force acting on the block. The vector sum of the vertical components of R and mg must thus cause this acceln, since they are the only forces with components in that direction. Hence we have:

mg-R\cos\theta = m a_v \ne 0 since a_v \ne 0

In short, you were assuming that the block was in vertical equilibrium, but it's not.
Ok thanks but in the question I was working on there was a horizontal force acting on the wedge itself. This force is just enough to keep the block from moving therefore the block has a vertical acceleration of 0. Could this just be a badly constructed question?
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dbs1984
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(Original post by TheAdviser101)
One thing that troubled me about this question was that if you split the reaction force into its real vertical component (r cos theta) and equate it to mg, the answer for the reaction force is wrong. Can you think of a possible reason why it does not work?
It is actually impossible to completely balance a block on a slope with only gravity and the reaction force. Gravity will pull straight down, but at any instant only part of the block will rest straight down on the slope. (As in, only part of the block is aligned in such a way so as to push downwards on the slope. The reaction is due to the part of the block resting on the slope pushing on it. Since only part of the block is resting on the slope, only part of the of gravity will push on it, and the reaction force is not gravity, but only a component of gravity: the one perpendicular to the slope. The component of gravity parallel to the slope will still pull on the block, and since there is no slope stopping the block from moving, it will move.

You said that there is a force, F acting on the block so that it does not move. Then the net upwards force on the block is not R cos ø, but R cos ø + the vertical component of F. This is what is equal to mg. And the horizontal force is then R sin ø - the horizontal component of F.

Your approach is correct, but it is very tedious. Instead of resolving each force into its horizontal and vertical components, which you will have to do for both R and F, you could resolve mg into a component parallel to the slope, and a component perpendicular to the slope. In that case, the component perpendicular to the slope will automatically be equal to R, and the component parallel to that slope will be equal to F. So instead of resolving two forces, you resolve one. To resolve gravity in such a way, construct a triangle, where one side is vertical (that's gravity, and it is your hypotenuse), one side parallel to the slope (which should be at an angle 90 - ø to the weight) and then draw a line perpendicular to the side parallel to the slope, and extend it until it meets the vertical line. Then use trig and find the components, keeping in mind that your hypotenuse is gravity
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atsruser
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(Original post by TheAdviser101)
Ok thanks but in the question I was working on there was a horizontal force acting on the wedge itself. This force is just enough to keep the block from moving therefore the block has a vertical acceleration of 0. Could this just be a badly constructed question?
I don't know - I haven't seen it. Put it up and I'll take a look.
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