A chain 72 meters long whose mass is 29 kilograms is hanging over the edge of a tall building and does not touch the ground. How
1 answer:
Answer:
Work done required is 3567.2 J
Explanation:
Given :
Length of chain, l = 72 m
Mass of chain, M = 29 kg
Linear mass density of chain, μ = =
= 0.40 kg/m
Let x be the length of the chain which lift to the top of the building.
Work done required to lift the chain is equal to the potential energy of the chain.
W = ∫μg (72 - x ) dx
Here g is acceleration due to gravity.
The limit of integration is from 0 to 14.
W = μg ( 72x - x²/2)
Substitute 0.40 kg/m for μ, 9.8 m/s² for g and 14 m for x in the above equation.
W =
W = 3567.2 J
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we can solve for the time :
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.