Question

A thick uniform rope of mass 6 kg and length 3 m is hanging vertically from a rigid support. The tension in the rope at a point 1 m from the support will be
(a) 20 N
(b) 30 N
(C) 40 N
(d) 60 N
plzzzz give solution not just the option plzzz​

Answer 1

Answer:

20 N

Explanation:

since it's a rope we will calculate the mass per unit length .

given mass m=6 kg

given length l= 3 m

therefore m/l= 2kg/m

hence the mass of the rope at 1m from the support will be m/l *1= 2kg/m *1m= 2kg

therefore the force of tension will be =mass*g=2*10=20N

plzzz like and rate my ans  : )

Answer 2

Given,

Total mass of rope = $6 kg$

This mass is entirely distributed throughout the length of the rope.

Then, mass per unit length of rope is

$M_l=\frac{Total\_mass\_of\_rope}{Total\_length\_of\_rope} \\\\M_l= \frac{6 Kg}{3 m} \\ \\M_l=2 \frac{Kg}{m}$

Tension at point $1m$ from the fixed end is

Tension = mass of rope below the point * acceleration due to gravity(g)

$T = M_{l(below\_1m)}*g$

$M_{l(below\_1m)} = M_l*(length\_of\_rope\_below\_1m)\\\\M_{l(below\_1m)} = 2\frac{Kg}{m} * 2m\\\\M_{l(below\_1m)} = 4 Kg$

Now, we know that value of $g = 10 \frac{m}{s^2}$

$T= (4Kg)*(10\frac{m}{s^2} )\\T = 40N$

∴Tension in the rope at point $1m$ below the fixed end is $T = 40 N$

Correct option is (c) 40 N

#SPJ2