What force had to act on that end?

Question

In the figure, the rock weighs 225 lb and is lifted 1 ft by the lever.

(a) How much work is done?

(b) The other end of the lever is pushed down 3 ft while lifting the rock. What force had to act on that end?

Answer 1

WORK FORMULA:

W =F*s*cosθ where

W = Work

F= Force

s= displacement (how much the object moved in a certain direction)

θ= the angle between the force and the direction of motion

FORCE FORMULA:

F=m*a where

m=mass

a=acceleration

WEIGHT FORMULA:

w=m*g where

w=weight

m=mass

g=gravity

In this problem a=g=32.2 ft/s²

PROBLEM (a):

W =F*s*cosθ

=m*g*s*cosθ where w=mg ----> m=w/g

=(w/g) *g*s*cosθ

=w*s*cosθ

=225lb*1ft*cos(0) θ= zero because the angle between the force and direction of motion is the same

=225 lb-ft

PROBLEM (b):

W =F*s*cosθ

F=W/[s*cos(θ)]

=225lb-ft/[3ft*cos(180)] θ= 180, the angle between the force and direction of motion is the opposite

=-75lb so the negative just means that you are pushed down 75 lbs