Weight of man on small wedge sliding on larger wedge

[Vector123]

Homework Statement

A man is standing on a small wedge that is sliding down a frictionless large wedge (base on the ground - horizontal) with angle 60. I don't have a picture so I'll just say that the surface of the smaller wedge is parallel to the ground. It has a weighing machine on it and the problem is to find the weight of the man.

Homework Equations

You could use a non-inertial frame attached to the smaller wedge carrying the weighing machine with inertial force gsin(60) pointing upward along the surface of the wedge and solve the problem easily...the answer is 0.25mg.

The Attempt at a Solution

My question is: The component of gsin(60) along the vertical is gsin(60)cos(30) which is 0.75g. So since the man's resting weight is mg and he is on a platform falling at 0.75g, his weight is 0.25 mg. Now the question is: Can you resolve components twice and add to get back the original vector? Seems to work in every case I tried...sorry if this is a silly question...I'm a just a hobbyist so I thought I would put the question in this section.

 

[Simon Bridge]

Sorry, you need a picture or a more careful description ... but this is usually best done using the non-inertial frame.

I'm guessing: there are two right wedges - a big one and a small one, each making similar triangles (same angles different sides).
The big one is hypotenuse upwards on the ground, and the small one is hypotenuse-down on top of it in such a way that the uppermost surface is horizontal.
The contact between them is frictionless.

A mass m sits on a sensitive balance on the small wedge as it slides down the big one.
Given the geometry of the wedges, what is the force measured by the balance.

This right?

Per your question: You can represent any vector as a sum of as many other vectors as you like. You may resolve the vector's components against any new set of axes as you like ad infinitiem and all the components will add up to the original one. It is easy to lose track though.

You can do the problem in an inertial frame using a free-body diagram - but care to identify all the forces involved.

 

[Vector123]

Sorry, you need a picture or a more careful description ... but this is usually best done using the non-inertial frame.

I'm guessing: there are two right wedges - a big one and a small one, each making similar triangles (same angles different sides).
The big one is hypotenuse upwards on the ground, and the small one is hypotenuse-down on top of it in such a way that the uppermost surface is horizontal.
The contact between them is frictionless.

A mass m sits on a sensitive balance on the small wedge as it slides down the big one.
Given the geometry of the wedges, what is the force measured by the balance.

This right?

Per your question: You can represent any vector as a sum of as many other vectors as you like. You may resolve the vector's components against any new set of axes as you like ad infinitiem and all the components will add up to the original one. It is easy to lose track though.

Thanks for your reply, Simon. Your description is spot on.

I managed to attach a picture to my original post but forgot to correct my post. If you would care to see it please look at the attachment to my post.

As you say, you can resolve a vector ad infinitum but in this particular case I was asking whether it is correct to resolve one of the components back again along the direction of the original vector it came from, so to speak. gsin(60) came from resolving g, so can gsin(60) be again resolved in the direction of g...I hope I've expressed my doubt somewhat clearly.

Thank you for your help; look forward to your answer.

 

[haruspex]

The component of gsin(60) along the vertical is gsin(60)cos(30) which is 0.75g. So since the man's resting weight is mg and he is on a platform falling at 0.75g, his weight is 0.25 mg.

You mean sin², not sin cos.

Now the question is: Can you resolve components twice and add to get back the original vector? Seems to work in every case I tried...sorry if this is a silly question...I'm a just a hobbyist so I thought I would put the question in this section.

Do you mean resolving twice in each direction, so that you get sin² and cos²? Certainly those will add back to the original.

 

[Vector123]

You mean sin², not sin cos.
Do you mean resolving twice in each direction, so that you get sin² and cos²? Certainly those will add back to the original.

Haruspex, thanks for your response. Since gsin(60) is the vector (hypotenuse) that is being resolved I was wondering whether it would be gsin(60)cos(30).

Hope I'm not making some very elementary mistake.

 

[haruspex]

Haruspex, thanks for your response. Since gsin(60) is the vector (hypotenuse) that is being resolved I was wondering whether it would be gsin(60)cos(30).

Hope I'm not making some very elementary mistake.

Sorry, I didn't notice the switch from 60 to 30. Yes, it's sin(60)cos(30), but it would be more instructive to use an unknown angle. Then it becomes $\sin(\theta)\cos(\pi/2-\theta)$, or more simply $\sin^2(\theta)$.

 

[Vector123]

Sorry, I didn't notice the switch from 60 to 30. Yes, it's sin(60)cos(30), but it would be more instructive to use an unknown angle. Then it becomes $\sin(\theta)\cos(\pi/2-\theta)$, or more simply $\sin^2(\theta)$.

Indeed it is, but if you resolve the other original component gcos(60) also you get gcos(60)sin(30).

So adding both vertical components (sorry, of the original components!) you end up with Sin(60+30) which is 1, so you get the original g back.

 

[haruspex]

Indeed it is, but if you resolve the other original component gcos(60) also you get gcos(60)sin(30).

So adding both vertical components (sorry, of the original components!) you end up with Sin(60+30) which is 1, so you get the original g back.

Yes, that's arriving at what I wrote in post #4, but by a different route:
$\sin(\theta)\cos(\pi/2-\theta)+\cos(\theta)\sin(\pi/2-\theta)=\sin(\theta+\pi/2-\theta)=\sin(\pi/2)=1$
$\sin(\theta)\cos(\pi/2-\theta)+\cos(\theta)\sin(\pi/2-\theta)=\sin^2(\theta)+\cos^2(\theta)=1$

 

[Vector123]

Yes, that's arriving at what I wrote in post #4, but by a different route:
$\sin(\theta)\cos(\pi/2-\theta)+\cos(\theta)\sin(\pi/2-\theta)=\sin(\theta+\pi/2-\theta)=\sin(\pi/2)=1$
$\sin(\theta)\cos(\pi/2-\theta)+\cos(\theta)\sin(\pi/2-\theta)=\sin^2(\theta)+\cos^2(\theta)=1$

Thanks Haruspex.

In my original post I was asking whether the answer I got from the inertial frame was correct because I resolved a component twice.

If you would kindly look at my original post my difficulty may become clearer. I obtained a weight of 0.25mg using both non-inertial and inertial frames. But the inertial frame calculation involves finding the component of a component.

Is that correct?

 

[Simon Bridge]

The way to approach your question is to keep everything as variables rather than use numbers. Then you can see if the situation works in general as opposed to just the examples you happen to have tried.

i.e. vertical could be $\vec v=v\hat\jmath$ (where $\hat\imath$ would be horizontal)
resolve into components in a coordinate system tilted by angle $\theta$ so $\vec v_x= v\hat x \sin\theta$ and $\vec v_y= v\hat y \cos\theta$

Notice that $\vec v = \vec v_x + \vec v_y$

But you can resolve the component vectors back into the original i-j coordinates... so try that.

 

[gracy]

Sorry to interrupt but

but if you resolve the other original component gcos(60) also you get gcos(60)sin(30).

For what?

 

[Vector123]

Sorry to interrupt but

For what?

That's exactly the point Gracy...it doesn't need to be resolved.

(My answer to haruspex was purely trigonometric, to show it added up to g)

Any comments on my original post? Am I making some elementary mistake in physics or trigonometry?

 

[Simon Bridge]

I think you are just not being careful in your notation and wording.
I think you have the tools to figure it out now.

 

[Vector123]

I think you are just not being careful in your notation and wording.
I think you have the tools to figure it out now.

Thanks for your interest, Simon, I appreciate it.

As I've said in my original post I solved the problem using a non-inertial frame.

But I was surprised to see the trigonometry of the forces and accelerations holding up in the intertial frame also, involving components of components, etc. For example, given gsin(theta) acceleration we can say that the horizontal acceleration is gsin(theta)cos(theta).

Assuming mgcos(theta) is canceled by the normal reaction of the larger wedge on the smaller one, the only force in the horizontal direction is mgsin(theta)cos(theta). In this case with theta=60 the acceleration works out to gsin(60)Cos(60) or sqrt(3)/4. One could use this to calculate the time taken to reach the halfway point from the top, etc, though there are other better ways to do it.

All this is true, of course, only if one can resolve a component of a vector along the original vector and cancel the normal reaction.

Anyway I was surprised at the result though no one else seems to be ! I just enjoy elementary physics, that's all.

Thanks again for your interest.