wedge

Since the system is actually resultantly accelerating downward( The block). Hence, the total downward force should be greater than the total upward force.

2010-06-15 14:21:36 補充：
In your question, the block is sliding down, right!?
And it is assumed to be no friction acting on the wedge by the block.
Rsinx and Rcosx are the component of the normal reaction acting on the wedge by the block.That's nothing wrong.
天同wrongly assume that the block is at rest.

2010-06-15 14:22:32 補充：
His follow calculator is right but just for the case of the block is at rest!

Your solution (the 3 red lines) is wrong.

There is, in fact, no need to resolve the weight of the block mg into two components. But if you do, the two components acting on the block is mg.sin(theta) and mg.cos(theta).

Since the block is at rest, mg.sin(theta) is balanced by friction Ff, which acts upward along the plane.
Thus Ff = mg.sin(theta)
Thus, the frictional force acting on the wedge = mg.sin(theta) which is downward along the plane (Newton's Third Law)'

Since the normal reaction acting on the block is mg.cos(theta), the force acting onto the wedge by the block = mg.cos(theta), which is perpendicular to the plane surface.

The resultant of these two forces
= square-root[(mg.sin(theta))^2 + mg.cos(theta))^2]
= mg.square-root[(sin(theta))^2 + (cos(theta))^2] = mg

Therefore, N = Mg + mg

In your solution:

N=Mg+Rcos(theta) <--- this is wrong, you need to include the force component given by friction.
The correct equation is,
N = Mg + R.cos(theta) + (mg.sin(theta)).sin(theta)
N = Mg + mg(cos(theta))^2 + mg.(sin(theta))^2
N = Mg + mg[(cos(theta)^2 + (sin(theta))^2]
N = Mg + mg

In fact, the answer of 115N is because of the use of a more exact value of g.

You use the gravitational acceleration,g is 10 m/s^2 but actually the gravitational acceleration is 9.81 m/s^2.

If you use g = 9.81 m/s^2,

N = 10 x 9.81 + 2 x 9.81 cos 30 = 115.09N