Question

A beam is placed evenly on a pivot point.On one side a 10N weight is placed 2 m from the pivot point and a 40N weight a further of 4 m from the pivot point.How far the pivot must the centre of gravity of an 80N,weight must be placed to perfectly balance the beam horizontally?

Answer 1

Explanation:

hope you will understand

Answer 2

Answer:

Beam is 3.25 m

Explanation:

• A beam is defined as element of the structure that will result the loads which is applied through lateral to beam’s axis direction. It is designed in such a way that it will carry loads of axial which would be column or strut in appearances.
• There, the load will be equal to result of a beam when the reaction force will result in support of beam.

Given that:

Beam that is placed on pivot point evenly.

One side = 10N

Weight = 2m

Other side =40 N

Weight = 4 m

The center of gravity of pivot = 80N

To find:

Weight of perfectly placed to balance the beam horizontally =?

Solution:

80x = 10x2 +40 *(2*4)

80x = 20+240

80x = 260

x= 260/80

x=13/4

x= 3.25 m

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