phy(mechanics)

用戶86611

2010-12-29 4:30 am

in the figure, a block of mass m rests on a wedge which in turn rests on a horizontal ground. the mass of the wedge is M and its sloping face makes an angle x with the horizontal. which of the following statements is/are correct?
(1) the normal force excerted by the wedge on the ground is(m+M)g
(2)there is friction acting on the wedge due to the ground
(3) the total force excerted by the block on the wedge is mgcos(x)

我唔明既就係(1)同(2),,因為我不明白何時個normal reacrion係(m+M)g 何時係Mg+mgcos(x)cox(x)??
同埋我都唔明第2點..

e.g ))a smooth block of mass 2 kg slides down a wedge. the wedge of mass 10 kg 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??

Ans:115N
the ans is calculated by Mg+mgcos(x)cox(x)=115N
why not(m+M)g ??

回答 (1)

天同

2010-12-30 12:16 am

✔ 最佳答案

As I have already pointed out in my other answer to you. The crucial point is whether there is friction present between the block and wedge.

If friction is present, there is an additional force of mg.sin(x) acting on the wedge, apart from the force mg.cos(x). The resultant of these two forces, by Pythagoras Theorem, is
= dquare-root[(mg.sin(x))^2 + (mg.cos(x))^2] = mg, and which is acting vertically downward.

If there is no friction present (between the block and wedge), there is only one force, mg.cos(x) acting perpendicular to the wedge surface. The verticl component of this force is thus equal to mg.(cos(x))^2

For your questions:Q1: The block is at rest on the wedge. This implies that there is friction between the block and wedge. The friction actong on the wedge is mg.sin(x). Hence, as explained above, the resultant force (of friction and normal reaction given by the block to wedge) acting on the wedge becomes mg.
Q2: Here, a smooth block is involved. The absence of frictional force between block and wedge leaves only a force of mg.cos(30) acting perpendicular to the wedge surface. By resolving component, it is not difficult to see that the vertically downward force is mg.cos(30).cos(30).Try to draw out free body diagrams of the wedge for the two different cases. You should be able to realize the differences.

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