Question

A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a second massless rope connected to a wall on the right at an angle of 130* from the horizontal rope (up 50.0* from the horizontal). What are the tensions in the ropes?g

Answer 1

The tension in the rope, T₁ = 205.6 N

The tension in the rope, T₂ = 319.8 N

Step-by-step explanation:

Given data,

The suspended mass, m = 25 kg

The angle with the horizontal, Ф = 50°

Let the vertical is the y-axis and horizontal is the x-axis,

Then, only T₂ has components in both x and y.

For equilibrium to exist, the magnitude of the component of T₂ in x, T₂ₓ, is equal to T₁,

The T₂y is equal to the force 'F' due to the 25 kg mass

The force F on mass, F = 245 N = T₂y.

T₂y = T₂ sin 50°

Therefore,

T₂ = T₂y / sin 50°

= 245 N / 0.766

= 319.8 N

Hence, the tension in the rope, T₂ = 319.8 N

T₁ = T₂ₓ

= T₂ cos 50°

= 319 x 0.6428

= 205.6 N

Hence, the tension in the rope, T₁ = 205.6 N

Answer 2

Answer:20.97g N,32.63g N

Step-by-step explanation:

We consider the forces at the knot.

The vertical forces are

`T_(2)Sin(50^(0)})` is the vertical component of tension
`T_(2)` at the knot.

`-25g` is the weight of the mass
`25Kg` acting downwards.

The horizontal forces are

`-T_(1)` is the tension in the rope acting left.

`T_(2)Cos(50^(0))` is the horizontal component of tension
`T_(2)` acting towards right.

Since the knot has no mass,it is always in equilibrium.

So,the sum of forces acting on it will be zero.

Balancing vertical forces gives,

`T_(2)Sin(50^(0)})`
`-25g=0`

`T_(2)`=
`32.63gN`

Balancing horizontal forces gives,

`T_(2)Cos(50^(0))`
`-T_(1)=0`

`T_(1)=32.63g* Cos(50^(0))=20.97gN`