A3.00-m-long ladder, weighing 200 n, leans against a smooth vertical wall with its base on ahorizontal rough floor, a distance of 1.00 m away from the wall. the ladder is not completelyuniform, so its center of gravity is 1.20 m from its base. what force of friction must the floor exerton the base of the ladder to prevent the ladder from sliding down

28.3 N

Explanation:

W = weight of the ladder = 200 N

L = length of the ladder = AC = 3 m

AD = distance of base of ladder from wall = 1 m

AB = distance of center of gravity from base = 1.2 m

F = force from the wall

f = frictional force acting between the ladder and floor

In triangle ACD

Cosθ =

Cosθ =

Sinθ =

Sinθ =

Sinθ =  

Using equilibrium of torque about the base of ladder

W Cosθ (AB) = F Sinθ (AC)

(200) () (1.2) = F () (3)

F = 28.3 N

Using equilibrium of force along the horizontal direction

f = F

f = 28.3 N

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.