✔ 最佳答案
Be aware that the question aks for the "normal reaction" exerted by the ground on the wedge. By "normal" reaction, it means the vertically upward force that the horizontal ground acts on the wedge.
The solution is apparent. Because the weights of both the block and wedge are acting vertically downward, by Newton's Third Law, the vertically upward reaction force (normal reaction) should be equal to the sum of the weights of block and wedge.
I don't know how you would get the expression Mg + mgcos^2(x). But the term mg.cos(x) is already not the "normal reaction" acting on the wedge.
2010-12-28 20:15:07 補充：
I got your point. The difference is that there is friction present in the first question. Hence, this gives rise to a force of mg.sin(x) acting downward along the inclined surface on the wedge. You need to find the vertical downward force of this component.
2010-12-28 20:18:07 補充：
The vertically downward force of mg.sin(x) is [mg.sin(x)].cos(90-x), which is equal to mg.sin^2(x). Combining this component with the one you have found (i.e. mg.cos^2(x)). You would get: mg.sin^2(x) + mg.cos^2(x) = mg, which is just the weight of the block.
2010-12-28 20:19:56 補充：
In your second question, because the block is smooth, friction is not present. Hence, the force component, mg.sin(x) is not present. This leaves only the component mg.cos^2(x).
