F4 physic force and motion 20分

alpha

2012-01-02 3:42 am

圖片參考：http://imgcld.yimg.com/8/n/HA00837988/o/701201010076013873417900.jpg

If the maximum friction between the block and the wedge is 4N
and the wedge accelerates towards the right,
what is the maximum acceleration of the wedge
so that the block remains relatively stationary on the wedge
1 唔明點解要搵個三角斜台既加速度?(個台會郁架咩?)
2 唔明點計.. 要計咩先
3 free body diagram
畫圖比我plz

回答 (1)

天同

2012-01-02 7:04 am

✔ 最佳答案

There are 3 forces acting on the 0.5 kg block
(i) the weight of the block 0.5g N (=5 N), acting vertically downwards
(ii) the normal reaction R, acting perpendicular upward to the wedge surface
(iii) friction 4 N, acting upward along the wedge surface

Because the block moves with the wedge with an acceleration a horizontally to the right (such that the block could remain almost stationary on the wedge), it doesn't have motion in the vertical direction, hence,
R.cos(30) + 4.sin(30) = 5
solve for R gives R = 3.464 N

For motion in the horizontal direction,
4.cos(30) - R.sin(30) = 0.5a
i.e. 4.cos(30) - 3.464sin(30) = 0.5a
solve for a gives a = 3.464 m/s2

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