Solving a Particle on an Inclined Plane: Help Needed!

[T.S.M]

Hey Everyone, here's a question i have issue with :(

A particle of mass 0.5 kg is pushed up a line af a greatest slope of a rough plane by a horizontal force of magnitude ℙ N. The plane is inclined at an angle 'α' (that is alpha) where tan α = 0.75 and the coefficient of friction between 'ℙ' and the plane is 0.5. The particle moves with constant speed. Find
a) the magnitude of the normal reaction between particle and the plane,
b) the value of ℙ.

I have been able to work out the normal reaction R , by R - (0.5g cos α) = 0.5 * 0. but then i recognized that ℙ is acting "horizontally" not parallel to the inclined plane, and so i wasn't able to solve it :( Secondly they mentioned that coefficient between 'ℙ' and the plane is .5 not between particle and plane! How's that possible :S pls explain it too! :/
Pls can any one help me out! Plz

 

[tiny-tim]

Hey T.S.M!

I have been able to work out the normal reaction R , by R - (0.5g cos α) = 0.5 * 0. but then i recognized that ℙ is acting "horizontally" not parallel to the inclined plane

Because the applied force is not parallel to the plane, the reaction force will depend on it …

resolve it into components along and perpendicular to the plane, and solve.

Secondly they mentioned that coefficient between 'ℙ' and the plane is .5 not between particle and plane! How's that possible :S

Must be a misprint

 

[T.S.M]

But reaction force can only be found out when we have the magnitude of P! And that's what we do't have! :S And the answer at the back of book is 9.8 N , both reaction force and P also! Both equal to 9.8 :/
Aaargh it's confusing! :O

 

[gneill]

But reaction force can only be found out when we have the magnitude of P! And that's what we do't have! :S And the answer at the back of book is 9.8 N , both reaction force and P also! Both equal to 9.8 :/
Aaargh it's confusing! :O

The particle is being pushed up the slope at constant speed. That should tell you something about the net sum of the forces acting along the slope.

 

[T.S.M]

The particle is being pushed up the slope at constant speed. That should tell you something about the net sum of the forces acting along the slope.

Hmm, i get what are you saying , but still 'P is UNKNOWN and R too' so 2 unknowns :P how can we solve it ;)
And may be it's misprinting or something, how can they both be equal to 9.8! :/
Umm btw u tried it? :P
Lemme know ASAP!

 

[tiny-tim]

how can they both be equal to 9.8! :/

Well, I make them equal …

get on with it! ​

 

[gneill]

Write the expressions for each force. Leave P as an unknown.

What's the component of P normal to the slope? Parallel to the slope?
What's the component of gravity on the particle normal to the slope? Parallel?

 

[T.S.M]

Write the expressions for each force. Leave P as an unknown.

What's the component of P normal to the slope? Parallel to the slope?
What's the component of gravity on the particle normal to the slope? Parallel?

For finding the normal reaction, R - (0.5g cos α) - (P sin α) = 05 * 0 (zero because, no acc. upwards)
Now equation for the motion of direction (P cos α) - (0.5g sin α) - (0.5* Value of R) = 0.5 * 0 (zero because, speed is 'constant')

So, am i going right? The magnitude of R which i m getting is in terms of P, and that's obvious! :O
Do you guys think my equations for both motions are correct?

 

[tiny-tim]

Yes, that's fine!

Now, tanα = 3/4, so sinα and cosα are … ?

 

[T.S.M]

Yes, that's fine!

Now, tanα = 3/4, so sinα and cosα are … ?

Aahhhhh! i got ittttttt!
•I made two equations, one equal to R, and other P, that is in both R and P were unknown and then solved it "simultaneously"

Husshhh!
Thank-you guys for Ur help, thanks alotttt! :D