Edexcel M2/M3 June 6th/10th 2013

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Right M2, if you have a ladder against a wall, or a plank on a diagonal resting on a peg or something, why is the reaction on the floor (usually A or something) vertially upwards??? and then if you have a horizontal plank with one end hinged at a wall and supported by a diagonal corner hinge plank thing, it is not vertical?

I know this is basic, but it confuses me

Reply 121

JoshThomas

Original post by JenniS

The reaction isnt actually upwards from the floor, that is just the normal reaction :smile:

its actual reaction is most probably at an angle, but by finding the normal reaction it makes resolving in a plane easier

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Reply 122

Boy_wonder_95

Original post by JenniS

The normal reaction is always perpendicular to the surface in contact :smile:

Reply 123

JenniS

Original post by JoshThomas

The reaction isnt actually upwards from the floor, that is just the normal reaction :smile:

its actual reaction is most probably at an angle, but by finding the normal reaction it makes resolving in a plane easier

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Original post by Boy_wonder_95

The normal reaction is always perpendicular to the surface in contact :smile:

Ok thanks, that explains the ladder ones. But what about the reaction on these poles hinged to a vertical wall held in equilibium by a rod joined to somewhere on the pole and the vertical wall......? (question 11 p148 if you have the textbook) why does the reaction not act in towards the pole?

Reply 124

Boy_wonder_95

Original post by JenniS

In that question, the reactions are at A and CD is basically just thrust, which is a pushing force.

Reply 125

JenniS

Original post by Boy_wonder_95

In that question, the reactions are at A and CD is basically just thrust, which is a pushing force.

ohhh god yeh, ah man I've been stupid and thinking about it wrong. cheers!!

Reply 126

Boy_wonder_95

Original post by JenniS

ohhh god yeh, ah man I've been stupid and thinking about it wrong. cheers!!

It's fine, we all make mistakes :wink:

Reply 127

currydud

does anyone know when to consider positive or negative work done? I just can't grasp it for example:
change in total energy= work done
what kind of scenario would you consider negative work done?

(edited 10 years ago)

Reply 128

MAD Phil

Original post by JenniS

First, let's look at bodies which are just touching each other, without a hinge. Then the normal reaction is at right angles to the line of contact between the bodies, and the friction (if there is any) is along the line of contact.

So you've got to think about the direction of the line where the two bodies touch. The end of a ladder is curved, so that doesn't help you know the direction of that line of contact. But if the end of the ladder is in contact with a horizontal floor, the surface of contact must be horizontal, and if it is resting against a vertical wall, it must be vertical. So the normal reaction must be vertical or horizontal, respectively.

But if a point somewhere along the length of the ladder is resting against a peg or the top of the wall, the other object is curved, and doesn't help you identify the line of contact. But the ladder is straight. So the line of contact is along the ladder. The normal reaction is at right-angles to the ladder and the friction (if then is any) is along the ladder.

Hinges are completely different. The whole point of their design (with a circle inside a circle) is that the normal reaction can point in any direction at all. So the hinge can provide any force that is needed to stop the two parts of the hinge parting company (while allowing the two parts to rotate relative to each other).

So when they tell you one body is hinged to another, you don't know the size or the direction of the force between them. You could put it on your diagram as an unknown force in an unknown direction, but usually we put it on as two unknown component forces in two directions at right-angles to each other. (They might be horizontal and vertical, or along a plane and at right-angles to it; whatever is convenient.) But those two forces are not normal reaction and friction.

Hope that helps, and good luck for next week.

Reply 129

MAD Phil

Original post by currydud

does anyone know when to consider positive or negative work done? I just can't grasp it for example:
change in total energy= work done
what kind of scenario would you consider negative work done?

Try post 119 above http://www.thestudentroom.co.uk/showpost.php?p=42892332&postcount=119

Any sort of resistance does negative work on the body, which reduces its energy. For instance friction or air resistance.

But other forces can also do negative work. Suppose your car won't start in the morning because it's got a flat battery. Your good friends push from behind to give you lots of kinetic energy so you can get the engine going. Your not-so-good friends push from the side, at 90 degrees to the direction of motion, so they look like they are helping but are really not doing any work. And your boyfriend/girlfriend who doesn't really want you to go to work yet pushes from the front. They are doing negative work, reducing the car's energy, so it doesn't start!

Reply 130

currydud

Original post by MAD Phil

thanks, that makes sense!
but why do they take away w.d rather than adding it on 7b? because wouldn't friction act the other way down the plane? so work done would be positive?
https://eiewebvip.edexcel.org.uk/Reports/Confidential%20Documents/0906/6678_01_que_20090522.pdf

Reply 131

MAD Phil

Original post by currydud

While the body is sliding up the plane, the friction opposes the motion by pointing down the plane. So it's doing negative work and causes the energy to decrease.

Once the body stops, there might have been enough friction to prevent it sliding back down again; that friction would have been pointing up the slope, but it wouldn't have done any work, because there wouldn't be any motion to help or oppose.

But in this question, there isn't enough friction to prevent the motion, so now the velocity is down the slope and the friction is up the slope. So again they are in opposite directions, the friction does negative work, and there is again a loss of energy.

In fact, once you see the word "friction" in an M2 or M3 question*, you know it can only cause a loss of energy, and will do so whenever the body slides.

Cheers, Ken

* It's possible for a moving body to gain energy because of friction with another moving body, but I'm pretty sure you won't get that in M2 or M3. Even then, the system of the two bodies together can still only lose energy because of the friction, and will do so whenever sliding occurs.

(edited 10 years ago)

Reply 132

currydud

Original post by MAD Phil

While the body is sliding up the plane, the friction opposes the motion by pointing down the plane. So it's doing negative work and causes the energy to decrease.

Once the body stops, there might have been enough friction to prevent it sliding back down again; that friction would have been pointing up the slope, but it wouldn't have done any work, because there wouldn't be any motion to help or oppose.

But in this question, there isn't enough friction to prevent the motion, so now the velocity is down the slope and the friction is up the slope. So again they are in opposite directions, the friction does negative work, and there is again a loss of energy.

In fact, once you see the word "friction", you know it can only cause a loss of energy, and will do so whenever the body slides.

thanks a lot!

Reply 133

JayJay95

Hey guys

Cab anyone explain 8c in the jan 2010 paper as i literally have no idea how the answer comes about. Same with 7b i dont know where the '6' comes from

Thanks :biggrin:

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Reply 134

Hamburglar

Original post by JayJay95

Hey guys

Cab anyone explain 8c in the jan 2010 paper as i literally have no idea how the answer comes about. Same with 7b i dont know where the '6' comes from

Thanks :biggrin:

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I second this, 8c just looks really weird and the mark scheme doesn't explain it well.

Thanks

Reply 135

Anonymous1717

Original post by Hamburglar

I second this, 8c just looks really weird and the mark scheme doesn't explain it well.

Thanks

So do I :/

Reply 136

Anonymous1717

Jan 2013 question 7b - why is e smaller than or equal to 1/2?
Thanks.

Reply 137

Boy_wonder_95

Original post by Anonymous1717

Jan 2013 question 7b - why is e smaller than or equal to 1/2?
Thanks.

Coefficient of restitution was 2e. It has to be less than 1.
2e < 1
e < 1/2

Reply 138

Anonymous1717

Original post by Boy_wonder_95

Coefficient of restitution was 2e. It has to be less than 1.
2e < 1
e < 1/2

That makes sense, but then is e not the coefficient of restitution in this equation? Coefficient of restitution cannot equal e if it's 2e.

Thanks a lot - I'm very grateful. :smile:

Reply 139