a uniform pole,PQ 30m long of mass 4kg is carried by a boy at p and a man 8m away from Q.find the distance from p where a mass of 20kg should be attached so that the man's support is twice that of the boy,if the system is in equilibrium.[g=10m/s~-2]
total mass is 24 kg (20 + 4)
... so man should be supporting 16 kg , and boy supporting 8 kg
working from p
... center of mass of PQ is 15 m
... man is supporting at 22 m
... 22 * support = 15 * 4 kg ... support = 60/22 kg
... 20 kg * d = (16 kg - 60/22 kg) * 22 m
i dnt understand pls explain it clearly
Explain further for mine understanding,please
To find the distance from point P where a mass of 20kg should be attached, we need to first understand the concept of moments in a system in equilibrium.
In a system in equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments.
In this case, the clockwise moment is provided by the man, and the anticlockwise moment is provided by the boy.
Let's solve the problem step by step:
1. Calculate the moment of the man:
Moment of the man = Weight of the man × Distance of the man from point P
Weight of the man = mass of the man × acceleration due to gravity
Weight of the man = 20kg × 10m/s^2
Weight of the man = 200N
Moment of the man = 200N × 8m
2. Calculate the moment of the boy:
Moment of the boy = Weight of the pole × Distance of the boy from point P
Weight of the pole = mass of the pole × acceleration due to gravity
Weight of the pole = 4kg × 10m/s^2
Weight of the pole = 40N
Moment of the boy = 40N × (30m - 8m)
3. Set up the equation for equilibrium:
Clockwise moment = Anticlockwise moment
Moment of the man = Moment of the boy
200N × 8m = 40N × (30m - 8m)
4. Solve the equation for the unknown distance:
1600m = 40N × 22m
1600m = 880m
Therefore, the distance from point P where a mass of 20kg should be attached so that the man's support is twice that of the boy is 22m.
So, the required distance from point P for attaching the mass of 20kg is 22m.