Question

A load 5kg is suspended from a rigid support A using a
uniform rope AC of length 2m and 4kg mass. The tension
in the rope at the point which is 0.5m below A is
(g=10ms)
1) 6ON 2) 8ON 3) 8N 4)4N
​

Answer 1

Answer:

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Explanation:

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Answer 2

Concept:

• Free body diagrams
• Net Forces
• Mass per unit length

Given:

• Length of rope = 2m
• Mass of rope = 4kg
• Mass per unit length of rope = 4/2 = 2kg/m
• Tension in the rope is at the point 0.5m below the top
• Mass of load = 5kg
• g= 10m/s^2

Find:

• Tension in the rope at point 0.5m below the top

Solution

The mass of the rope below the point of tension = 1.5*2 = 3kg

Total mass = mass of rope + mass of load = 3+5 = 8kg

Total tension = mg = 8*10 = 80N

The tension in the rope at the point which is 0.5m below A is 80N.

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