Question

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the beam to be balanced, a 2.9-kg mass must be placed on the left-end, and a 8.00-kg mass must be placed at another location on the beam. How far from the left-end is the location of the 8.00-kg mass

Answer 1

x ’= 1,735 m, measured from the far left

Step-by-step explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

x_{cm} = 1.2 -1

x_ {cm} = 0.2 m

Σ τ = 0

w₁ 1.2 + mg 0.2 - W₂ x = 0

x =
`(m_1 g\ 1.2 \ + m g \ 0.2)/(M_2 g)`

x =
`(m_1 \ 1.2 \ + m \ 0.2 )/(M_2)`

let's calculate

x =
`(2.9 \ 1.2 \ + 4 \ 0.2 )/(8.00)`2.9 1.2 + 4 0.2 / 8

x = 0.535 m

measured from the pivot point

measured from the far left is

x’= 1,2 + x

x'= 1.2 + 0.535

x ’= 1,735 m