Question

A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizontal. What is the horizontal force exerted by the ground on the ladder when an 81 kg object is 4.0 m from the bottom?

Answer 1

266.5 newton

Step-by-step explanation:

Weight of first object = 51 kg * 9.81

= 500 N

Weight of second object = 81 kg *9.81

= 794.61 N

As length of ladder is 15 m, the center of mass will be at 7.5 m.

Level arm is always perpendicular to the force. So the level of arm is at

7.5 sin θ.

τ ( wall) = τ (first object ) + τ ( second object)

N cos 30 ×15 = 794.61 sin 30 × 4 + ( 500 sin 30 ×7.5 )

13 N = 1589.22 + 1875

13 N = 3464.22

N= 266.5 newton

Answer 2

Step-by-step explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N