In order to lift a lighter object on the other side, a boy placed 155 N of effort force through a vertical distance of 0.25 m. How far will the lever lift
the lighter object if the resistance distance is 3.5 m from the fulcrum and
the effort distance is 1.5 m from the fulcrum? *

0.58 m

Explanation:

In order to find the vertical distance the load lifts, we first need to find the magnitude of the resistance force. So, Rd = Fd' where R = resistance force = ?, d = resistance distance = 3.5 m, F = effort force = 155 N and d' = effort distance = 1.5 m.

So R = Fd'/d = 155 N × 1.5 m/3.5 m = 232.5 Nm/3.5 m = 66.43 N  

Now, the vertical work done by the effort = vertical work done by load

Fy = Ry' where y = vertical distance moved by effort = 0.25 m and y' = vertical distance moved by resistance force

So, Fy = Ry'

y' = Fy/R

= 155 N 0.25 m/66.43 N

= 38.75 Nm/66.43 N

= 0.58 m

So, the vertical distance the lighter object moves is y' = 0.58 m

this is true

explanation:

i looked it up

answer:

fvbyunjk

explanation: