A3.00-m-long ladder, weighing 200 n, rests against a smooth vertical wall with its base on a horizontal rough floor, a distance of 1.00 m away from the wall. if the center of mass of the ladder is 1.20 m from its base, what frictional force must the floor exert on the base of the ladder in order for the ladder to be in static equilibrium?
a) 102 n
b) 93.3 n
c) 28.3 n
d) 150 n
e) 130 n

The frictional force must be 28.3 N ( option C)

Explanation:

Step 1: Data given

Length of the ladder = 3.00 meter

The ladder eighs 200 N

Distance away from the wall = 1.00 meter

Center of mass of the ladder is 1.20 m from its base

Step 2: Calculate the height of the ladder

Let's consider the space between the ladder and the wall as a right triangle.

Pythagoras says:

Base² + Height² = Hypotenuse²

with base = 1.00 meter

with height = TO BE DETERMINED

with hypotenuse = 3.00 meter

1²+ h² = 3²

1 + h² = 9

h = √8

Step 3: Calcuate  the angle between ladder and floor

Consider the angle θ = the angle the ladder makes with the floor

Sin θ = h/L = √8/3

sin θ = 0.9428

cos θ = 1/3

θ ≈ 70.5°

Step 4: Calculate torque

Since the center of mass is 1.20 m from the ladder’s base,

the torque, caused by the ladder’s weight, can be calculated as followed:

dh = 1.2 * cosθ = 1.2 * 1/3 = 0.4m

Torque = 200N * 0.4m = 80

This is the counter clockwise torque.

Step 5: Calculate frictional force

The force from the wall  on the top of the ladder, is one of the horizontal forces. The friction force is the other horizontal force. These two forces are equal.

The vertical distance from the wall force to the floor is equal to the height of the right triangle.

Torque = F * h

F = torque /h

F = 80 ÷ √8

F = 28.28 ≈ 28.3 N

The frictional force must be 28.3 N ( option C)

coil wire, increase current, put iron inside the coil

i guess it’s a and b