What force must the man apply to one end of the crowbar?
A 6 ft. crowbar is employed to end a 400 lb load. A stone(acting as a fulcrum) is placed under the crowbar 8 inches away from the end on which the shipment rests. Neglect the weight of the crowbar.
400lbs-->Newtons = 4000N
8 inches---> meters = 0.2032 meters
Jiffy of Force = equal to mass(N) X perpedincular distance(m)
F= 4000N X 0.2032 meters
= 812.8 nm
haughtiness on other side --> 8 inches - 6 feet = 1.6256 meters
Moment of Duress on one side is always equal to moment of force on the other side
Moment 1 = Moment 2
812.8nm= f*1.6256 meters
812.8nm/1.6256 meters= F
500=N
The ans is 500N
8 inches---> meters = 0.2032 meters
Twinkling of an eye of Force = equal to mass(N) X perpedincular distance(m)
F= 4000N X 0.2032 meters
= 812.8 nm
rigidity on other side --> 8 inches - 6 feet = 1.6256 meters
Moment of Power on one side is always equal to moment of force on the other side
Moment 1 = Moment 2
812.8nm= f*1.6256 meters
812.8nm/1.6256 meters= F
500=N
The ans is 500N






