Question

a 15m ladder weight 50kg rest against smooth wall at a point of 12m above the ground. the centre of gravity ladder is one third the way up .A 80kg man climb the half way up the ladder . find the force exert by the system on the ground and wall?

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Answer 1

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$\large \underline \orange{Qᴜᴇsᴛɪᴏɴ:- ☻}$

A 15m ladder weight 50kg rest against smooth wall at a point of 12m above the ground. the centre of gravity ladder is one third the way up .A 80kg man climb the half way up the ladder . find the force exert by the system on the ground and wall?

$\large \underline \red{ᴀɴsᴡᴇʀ:- ☻}$

Suppose, The angle made by the ladder with ground be angle θ.

⇨cosθ=( 12/15 )

⇨cosθ=( 3/5 )

So,

⇨sinθ= 4/5

The weight of the object will be concentrated at the point at the distance of 5 meter from the ground and the man is at a distance of 7.5 meter from the ground.

The net force acting on the ladder should be equal to zero because the ladder is resting on the ground.

The moment of all the forces about the point where the ladder is resting we get :

⇨S(15sinθ)=mg(5cosθ)+Mg(7.5cosθ)

⇨S(15)( 4/5 )=500(5) 3/5 +800(7.5) 3/5

⇨12ꜱ=(1500+3600) N

⇨12ꜱ=5100N

⇨ꜱ=425N

As the net vertical force is also zero in order to be in equilibrium,

ʀ=ᴍɢ+ᴍɢ

ʀ=(500+800)ɴ

ʀ = 1300 ɴᴇᴡᴛᴏɴ

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Answer 2

Answer:

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Qᴜᴇsᴛɪᴏɴ:−☻

A 15m ladder weight 50kg rest against smooth wall at a point of 12m above the ground. the centre of gravity ladder is one third the way up .A 80kg man climb the half way up the ladder . find the force exert by the system on the ground and wall?

ᴀɴsᴡᴇʀ:−☻

Suppose, The angle made by the ladder with ground be angle θ.

⇨cosθ=( 12/15 )

⇨cosθ=( 3/5 )

So,

⇨sinθ= 4/5

The weight of the object will be concentrated at the point at the distance of 5 meter from the ground and the man is at a distance of 7.5 meter from the ground.

The net force acting on the ladder should be equal to zero because the ladder is resting on the ground.

The moment of all the forces about the point where the ladder is resting we get :

⇨S(15sinθ)=mg(5cosθ)+Mg(7.5cosθ)

⇨S(15)( 4/5 )=500(5) 3/5 +800(7.5) 3/5

⇨12ꜱ=(1500+3600) N

⇨12ꜱ=5100N

⇨ꜱ=425N

As the net vertical force is also zero in order to be in equilibrium,

ʀ=ᴍɢ+ᴍɢ

ʀ=(500+800)ɴ

ʀ = 1300 ɴᴇᴡᴛᴏɴ

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