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**pokrate****Member**- Registered: 2012-11-16
- Posts: 13

Hi,

This is my first post, so bear with me please.

I was solving normal friction less inclined plane problem and was trying to find velocity at the bottom of the plane.

All texts focus on using co-ordinate system with X axis along the plane, and 0,0 at the position of the block lying at the top, and Y axis perpendicular to the inclined plane.

I was trying to use 0,0 as the bottom of the inclined plane, and I get a = g/sin(theta) which means a = infinite for theta = 0.

Please help me !!!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,912

hi pokrate,

Welcome to the forum.

I've made you a fresh diagram (see below).

I've put the acceleration due to gravity in red.

My diagram software won't let me do theta so I've used 'a' for the angle of the slope.

The dotted lines show directions that are parallel and perpendicular to the slope.

You then have to split **g** into two components; one parallel to the slope and one perpendicular to the slope.

Notice that the angle a is repeated in the red 'acceleration' triangle.

So the component you want is **g sin(a)**.

Hope that helps.

Bob

ps. I've never found I needed coordinates at all for problems like these. Only time I've used them is to show that an object projected at an angle, and flying through the air, follows a parabolic path (if you ignore wind resistence).

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**pokrate****Member**- Registered: 2012-11-16
- Posts: 13

bob bundy wrote:

hi pokrate,

Welcome to the forum.

I've made you a fresh diagram (see below).

I've put the acceleration due to gravity in red.

My diagram software won't let me do theta so I've used 'a' for the angle of the slope.

The dotted lines show directions that are parallel and perpendicular to the slope.

You then have to split

ginto two components; one parallel to the slope and one perpendicular to the slope.Notice that the angle a is repeated in the red 'acceleration' triangle.

So the component you want is

g sin(a).Hope that helps.

Bob

ps. I've never found I needed coordinates at all for problems like these. Only time I've used them is to show that an object projected at an angle, and flying through the air, follows a parabolic path (if you ignore wind resistence).

If I read my textbooks first and then try to solve such problems, I can solve like you. But if I try to take a look at this problem from a fresh perspective, then I am still looking for a meaningful answer.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,912

My approach is to draw a diagram first. Sometimes I find lines get in the way so it may take a few re-draws before I'm happy.

Then I put on all the forces (rather than accelerations) because when you have friction as well you'll need F and the normal reaction, R, and these are forces. If I'm helping someone on-line, I stick to {diagram in black or dark blue}, and {forces in red} so they stand out.

Then I resolve parallel and perpendicular to the slope to get the equations.

I also write down what I else know, like u = 0, s (for distance) or t (for time) = whatever, acceleration = gsin(a) and what I want to know; in your case v.

Then I can choose the right formula; v = u + at or v^2 = u^2 + 2as, depending on the rest of the question.

If you get stuck on one, you are welcome to post it here and I'll try to give some hints.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**pokrate****Member**- Registered: 2012-11-16
- Posts: 13

bob bundy wrote:

My approach is to draw a diagram first. Sometimes I find lines get in the way so it may take a few re-draws before I'm happy.

Then I put on all the forces (rather than accelerations) because when you have friction as well you'll need F and the normal reaction, R, and these are forces. If I'm helping someone on-line, I stick to {diagram in black or dark blue}, and {forces in red} so they stand out.

Then I resolve parallel and perpendicular to the slope to get the equations.

I also write down what I else know, like u = 0, s (for distance) or t (for time) = whatever, acceleration = gsin(a) and what I want to know; in your case v.

Then I can choose the right formula; v = u + at or v^2 = u^2 + 2as, depending on the rest of the question.

If you get stuck on one, you are welcome to post it here and I'll try to give some hints.

Bob

I guess you haven't taken a look at the figure I posted. Using that diagram I just want to see how a = g.sin(theta) ? Because using the frame I posted it is easy to see that a = g / sin(theta) and not g.sin(theta).

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,919

Hi pokrate

The acceleration of the object from the problem is the consequence of the gravitational acceleration. The gravitational acceleration can be split into two accelerations, one perpendicular to the inclined plane and one parallel to it, both less than g itself. The perpendicular part doesn't have any effect on the object, while the parallel part is the acceleration you want. The two are represented in Bob's picture, from which we see that a=g*sin(theta).

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,912

hi pokrate,

I assure you that I did look carefully at your diagram.

Please think about which is the hypotenuse, and which, the opposite. You have them the wrong way round in your mind.

The acceleration due to gravity must split into two components. Each component must be less than g or you are creating energy from nothing.

I assure you my diagram is the correct one. I have taught this topic successfully for 40 years.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**pokrate****Member**- Registered: 2012-11-16
- Posts: 13

anonimnystefy wrote:

Hi pokrate

The acceleration of the object from the problem is the consequence of the gravitational acceleration. The gravitational acceleration can be split into two accelerations, one perpendicular to the inclined plane and one parallel to it, both less than g itself. The perpendicular part doesn't have any effect on the object, while the parallel part is the acceleration you want. The two are represented in Bob's picture, from which we see that a=g*sin(theta).

I like the explanation, as if something can be split into a direction, then it must be able to split it into other direction which is not possible in my case. So, that is the reason why a = g.sin(theta). Thanks for the simple explanation.

Now to wet some appetite :

Attached is a figure showing grooved inclined plane, of which side view will be like the figure we are discussing and also attached in my first post. AC = 4, BC = 3, and the angle of the grooved path is @, now if the ball is released from rest from the top of the plane, what will be its velocity at point B.

*Last edited by pokrate (2012-11-17 07:00:22)*

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