What are the forces acting on a particle in equilibrium on an inclined plane?

[jernobyl]

Homework Statement

This is from exercises on Coplanar Forces and Moments.

A particle of mass 10kg rests in equilibrium on a smooth plane inclined at 30 degrees to the horizontal, being held by a light string inclined at 15 degrees to the line of greatest slope of the plane. Find the tension in the string and the magnitude of the reaction of the plane on the particle.

Homework Equations

Basic resolving forces...so basic I can't even do it.

The Attempt at a Solution

Well, I have resolved the 10g force to be parallel to the slope:

10 x 0 (a=0 because particle is in equilibrium?) = T - 10g.sin30
T = 10g.sin30
T = 49

and then I wondered if all I had to do was now resolve this force to be 15 degrees "up" as such but when I do that:

T = 49cos15
T = 47.33 N

Aaaaaaaand apparently the answer is meant to be 50.76N.

As for finding the reaction force...no idea. Any help with that would be well received.

Many thanks!

 

[Dick]

Well, it's not the easiest problem in the world. But you have three basic forces. You have the known weight of the block pointing straight down at 10kg*g. You have an unknown normal force pointing perpendicular to the plane. And you have an unknown tension pointing opposite the downward direction of the plane at 45 degrees to the horizontal, right? Ok, so all of the directions are known, but the magnitude of two of the forces is unknown. Nevertheless, the cancellation of the horizontal and vertical components of each of the forces should give you two equations in two unknowns. Just try it and be patient. The problem is a bit off the beaten track but at least everything is coplanar. Whew!

 

[HallsofIvy]

This is a physics problem rather than math so I am moving it.

 

[jernobyl]

Thanks for your reply, Dick, but err...I'm not entirely sure what you're saying. The "cancellation" of the horizontal and vertical components? I'm not following...

 

[jernobyl]

Hello? Anyone? :(

 

[Dick]

You can split all of the forces either horizontally and vertically or normal and parallel to the plane of the incline. Neither is easier than the other. But in the end all of the forces have to cancel since the object is not accelerating. This will give you two equations in two unknowns. Just try it and post your results and we can help from there.

 

[jernobyl]

Okay, well, erm, I've done it parallel and normal to the plane.

Resolving along the plane:

0 + Tcos15 - 10g.sin30 = 0

And perpendicular to the plane:

N + Tcos15 - 10g.cos30 = 0

I don't even know if that's right or not. I wish I just understood.