phy (mechanics)

用戶86611

2010-12-29 2:15 am

a wedge of mass M is fixed to the ground. a block of mass m slides down the wedge with constant velocity as shown. what is the normal reaction exerted by the ground on the wedge??

ans:(M+m)g

P.S 我想問點解個 normal reaction 不曷Mg+mgcos^2x???
有咩情況下個normal reaction係Mg+mgcos^2x??關唔關個wedge fixed to the ground事??同埋關唔關佢係at rest or constant vel.事??因為有d問題係個normal reaction係Mg+mgcos^2x......

圖片參考：http://imgcld.yimg.com/8/n/HA01002325/o/701012280118913873432710.jpg

更新1:

for example: a smooth block of mass 2kg slides down a wedge. the wedge of mass 10kg, is placed on a horizontal table, and its inclined plane makes an angle of 30 with the horizontal. if the wedge remains stationary all the time, the normal reaction of the table acting on the wedge is??

更新2:

ans:115N and the ans calculated is used Mg+mgcos(x)cos(x) which is not equal to (M+m)g

回答 (1)

天同

2010-12-29 3:43 am

✔ 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).

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