A uniform rod PQ of length 1m and mass 2kg is pivoted at the end p. If load of 14N is placed at the center of the rod, find the force that should be applied vertically upwards at Q to maintain the rod at equilibrium horizontally [g = 10ms^-2]

Brinsage23

5 years ago

Since the rod is 2kg (20N) and a load of 14N is place at the centre. The total load on the centre of the rod will be 34N (20N + 14N = 34N)
According to the principle of moment:
Clockwise moment = Anticlockwise moment
34(0.5)= F × 1
17N = F
F = 17N

Mudizzz

6 years ago

clockwise=anticlockwise
the solving is self explanatory

Onesimus001

5 months ago

Can i still generate the same profile code with another number

tirimisiyu

10 years ago

can u pls solve cos I don't understand

missie

10 years ago

pls solve it

Daniel Brainy

9 years ago

At equilibrum,

upward moment is equal to downward moment. The force to be applied vertically upward multiplied by its distance from d pivot 1m is equal to product of the sum of the rod weight 20 N and the force at d centre 14N and their common distance 0.5m from d pivot.