14.A metre rule is pivoted at its mid point vertically. Downward
forces of 2N and 6N are applied at the 10 cm and 80 cm marks
respectively. What vertical upward force must be applied at the 90
cm mark if the rule is to be balanced horizontally.
Answers
Answered by
3
Answer:
2.5 N
Explanation:
A METRE RULE HAS A LENGTH OF 100 CM
Torque at 10 cm = F x OP
= 2N * 40/100 m
= 40/50 N m = 0.80 N m
Torque at 80 cm= F x OP
= 6N * 30 cm
= 6N * 30/100 m
= 3*3/5 N m
= 1.80 N m
To balance the rule, Torque on LHS must be = to Torque on RHS.
So, At 90 cm, a torque of -1 N m (1 N mbecause it is applied upwards) should be applied.
Torque at 90 cm = F x OP
1 = F x 40/100 m
1 = F x 0.4
1/0.4 = F
F = 2.5 (N) = 2.5 N
Hence, a force of 2.5 N must be applied vertically upwards to balance the rule horizontally.
Similar questions