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

Explanation:

We consider the forces at the knot.

The vertical forces are

is the vertical component of tension at the knot.

is the weight of the mass acting downwards.

The horizontal forces are

is the tension in the rope acting left.

is the horizontal component of tension 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,

=

Balancing horizontal forces gives,

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

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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s