ICSE Class 10 Answered
the main gate of the building is 2.6m broad.it can be opened by applying a force of 80N at the middle of the gate
1) calculate the least force which can open the door
2) where should the force be applied
Asked by pradipdhole | 10 Jun, 2019, 10:57: PM
Expert Answer
The force applied at the middle of gate = 80 N
The distance of gate from hinge = 2.6 m
The distance at which the 80 N of force is applied = 2.6/2 = 1.3 m
The minimum force will be required if the force is applied at the free end of the gate = 2.6 m
Thus, by principle of moments
F1 x d1 = F2 x d2
80 x 1.3 = F2 x 2.6
Thus, F2 = (80 x 1.3)/2.6 = 40 N
1) Thus, least force which can open the door is 40 N
2) The force has to be applied at the distance of 2.6 m from the hinge of the gate to open the gate with least force.
Answered by Shiwani Sawant | 11 Jun, 2019, 12:51: PM
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