Asked by Eli
It is found out that a uniform beam 100cm long and of mass 95g can be balanced on a knife-edge when a 5g mass is hung 10cm from one end. How far is the knife-edge from the centre of the beam?
Answers
Answered by
GPT 3.5
Let x be the distance from the center of the beam to the knife-edge.
The total mass of the beam is 95g, the mass hanging on one side is 5g, and the remaining mass on the other side is 95g - 5g = 90g.
Let the distance from the center of the beam to the 5g mass be 10cm. The distance from the center to the center of mass of the remaining 90g is x cm.
To balance the beam:
(5g)(10cm) = (90g)(x)
50 = 90x
x = 50/90
x = 5/9
Therefore, the knife-edge is located 5/9 cm from the center of the beam.
The total mass of the beam is 95g, the mass hanging on one side is 5g, and the remaining mass on the other side is 95g - 5g = 90g.
Let the distance from the center of the beam to the 5g mass be 10cm. The distance from the center to the center of mass of the remaining 90g is x cm.
To balance the beam:
(5g)(10cm) = (90g)(x)
50 = 90x
x = 50/90
x = 5/9
Therefore, the knife-edge is located 5/9 cm from the center of the beam.
There are no human answers yet. A form for humans to post answers is coming very soon!