Physics, 22.11.2019 19:31 Jenan25

A25kg 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

A25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the le

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Answer from: sweetmochi13

20.97g N,32.63g N

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,

$T_{2}Cos(50^{0})$ $-T_{1}=0$

$T_{1}=32.63g\times Cos(50^{0})=20.97gN$

Answer from: laytonlutz

The tension in the rope, T₁ = 205.6 N

The tension in the rope, T₂ = 319.8 N

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 from: Quest

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Answer from: Quest

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