A uniform plank of weight 450 N rests on two trestles X and Y and a worker of weight 675 N stands at one end of the plank. The worker holds a light basket which contains several packets of goods each of weight 6 N. What is the maximum number of packets he can hold without tilting the plank?
(Assume the weight of the plank acts through its centre.)
why shouldn't we concern about the supporting force from trestle X?
how to determine whether we should consider the normal reaction when in other situations?
So in the above case, the two forces balance each other? However, how could we know that they are balanced?
- 翻雷滾天 風卷殘雲Lv 71 十年前最愛解答
Taking moment about Y, we have:
Anti-clockwise moment = 450 x 2.5 = 1125 N m
So at/before critical point, the clockwise moment about Y due to the weights of the worker + goods should be smaller than or equal to 1125 N m, i.e.
(675 + 6n) x 1.5 <= 1125 where n is the no. of packets
675 + 6n <= 750
n <= 12.5
So the max. no. of packets he can hold is 12
2010-12-23 17:01:57 補充：
Becos at the critical point, the plank just loses contact with X and so the normal reaction from X will be zero.
2010-12-25 19:55:04 補充：
When any two moments about the same point are NOT balanced with each other, there must be 3rd (or even more) moment(s) about the same point that we need to consider.
2010-12-25 22:20:34 補充：
In the above case, there are 3 forces (NOT moments):
1) Weight of plank
2) Weight of man + basket
3) Normal reaction at Y
At critical point, (3) = (1) + (2) (when considering forces)
But when talking about moments, only (1) and (2) contribute when taking about Y.資料來源： 原創答案