Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

I'm abusing your help

A body of mass 2kg is on the point of slipping up on a rough plane inclined at 20º to the horizontal, and is being acted upon by a force Y at an angle of 45º to the horizontal. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle, find Y.

I get 25.14N. But I'm wrong.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: I'm abusing your help

There are four directions Y can point without being specific.
There are two directions the inclined plane can point without being specific.
Please clarify, and also include a list of all forces by name with a variable name.

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igloo myrtilles fourmis

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: I'm abusing your help

Ok here's the diagram. The numbers in black are what I've been told in the question, and the numbers in red are what I have added in myself.

To find Y:

The book says Y = 43N. What have I done wrong?

Thanks.

EDIT: I'm not really sure I've put the 45º angle in the right place.. but that's the only place I could think to put it.

Last edited by Daniel123 (2007-10-25 09:14:05)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: I'm abusing your help

I'd have put Y pointing towards the top-right of the page. I only even thought of putting it your way when John started complaining.

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Why did the vector cross the road?
It wanted to be normal.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: I'm abusing your help

Hmm...

I still don't see how to get to that answer though? The question isn't worded very well.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: I'm abusing your help

The magnitude of the normal reaction is not equal to the weight of the body in this problem! Don’t always assume this to be the case in every problem!

Let R be the normal reaction. Then resolve forces along and at right angles to the inclined surface:

Along the surface: 19.6sin20° + 0.2R = Ycos65°

Perpendicular to it: 19.6cos20° + Ycos25° = R

Now substitute the value of R from the second equation into the first equation and you can then solve for Y.

Last edited by JaneFairfax (2007-10-26 02:57:58)

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: I'm abusing your help

I wasn't thinking.. clearly if the particle is on an inclined plane then the reaction caused by the weight of the particle will be equal to the component perpendicular to the plane, and not the actual magnitude of the weight itself. If there is an external force Y, unless it acts along the same line as the plane, then it too will contribute to the reaction, as it is increasing the force exerted by the particle on the plane.

I understand now, and I can get to the correct answer. This also explains why my answers to the previous questions were all a bit off... I was just assuming that R=W.

Thank you(pl) once again