A uniform meter rule weighing 0.5N is to be pivoted on a knife-edge at the 30cm-mark. where will a force of 2N be placed from the pivot to balance the meter rule?
95cm
25cm
20cm
5cm
Explanation
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when d meter rule was pivoted @ d 30cm mark,0.5N was suspended meaning 20cm from the COG(pivoted at 50cm).At d other end,2N was suspended so d length from d center is x
with simple calculations it gives 5.
The load of 2N will be placed 5cm away from the pivot which is at the 30cm mark but the load is placed at the 25cm mark. Because 30-5(where 5 is distance away from the pivot) equals 25.
The weight of the meter rule is at the centre marked 50cm, hence, 20cm from the pivot
2N weight should be placed d cm from the other end
Therefore, 2 x d = 0.5 x 20
d = 5cm
Ans = D
if the 2N force where placed at the 5cm mark then the distance from the pivot would be 25cm
it is 25cm since to calculate for 2N we are going to minus the distance from the pivot
answer cannot be d ooo
cos the lower d distance from the pivot den it will bend toward d greater force since d lower force is close to d pivot
