A30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force →f so that the swing ropes are 30.0° with respect to the vertical. (a) calculate the tension in each of the two ropes supporting the swing under these conditions. (b) calculate the magnitude of →f .

The tension in each of the two ropes supporting the girl, swing under these condition is 170 N and the magnitude of is 170 N.

Tension is the pulling force carried by the flexible mediums like ropes, cables, and string. Tension in a body due to the weight of the suspension body is the net force acting on the body.

For the body with mass (m) hanging on a string, the tension force of the string for the component of weight in the direction of sting, can be given as,

Here, (g) is the gravitational force.

The weight of the girl is 30.0-kg. The girl held at rest by a horizontal force, so that the swing ropes are 30.0° with respect to the vertical. Thus, put the values in the above formula as,

Thus, the tension in each of the two ropes supporting the swing under these conditions is 170 N.

The tension force of the string, for the component of weight perpendicular to the sting, can be given as,

In the terms of Force,

Put the values,

Thus, the magnitude of is 170 N.

The tension in each of the two ropes supporting the girl, swing under these condition is 170 N and the magnitude of is 170 N.

Learn more about the tension in the string here;

link

Part a)

Part b)

F = 170 N

Explanation:

As we know that swing is at rest in this situation

so here all the forces on the swing must be balanced

so here we have

so here we have

Now in the horizontal direction force is also balanced

so we have

now we have

I will answer in English.

Ok, we know that a ball is trowed up, and the maximum height is 21.4m

If we suppose that the initial height of the ball is h = 0 (this means that is trowed from the ground) the equations of movement will be:

for the acceleration we only have the gravitational acceleration, so we have:

a(t) = -9.8m/s^2

for the velocity we can integrate over time and get:

v(t) =  (-9.8m/s^2)*t + v0

where v0 is the initial speed.

For the position we integrate over the time again, and as the initial position is 0m, here we do not have any integration constant.

p(t) = (1/2)(-9.8m/s^2)*t^2 + v0*t

a) the maximum height is reached when the velocity is equal to zero, this happens at the time:

(-9.8m/s^2)*t + v0 = 0

t = v0/(9.8m/s^2)

now we can replace this time in the equation for the position:

p = 21.4m = (-4.9m/s^2)*(v0/(9.8m/s^2))^2 + v0^2/(9.8m/s^2)

and solve it for v0, i will stop writing the units so it is easier to read:

21.4 = -4.9*(v0/9.8)^2 + v0^2/9.8

21.4 = v0^2*(0.05)

v0 = √(21.4/0.05) = 20.7

so the initial speed is 20.7 m/s.

b) the things i supposed are:

The initial position is p(0) = 0

there are no things like air resistance or wind.

c) we can take t = 3.05s and put this in the position equation:

p(3.05s) =  (1/2)(-9.8m/s^2)*(3.05s)^2 + 20.7m/s*3.05s = 17.5m

3.05 seconds after the launch, the ball will be 17.5 m above the ground.

d) we put t = 3.05s in the velocity equation:

v(3.05s) =  (-9.8m/s^2)*3.05s + 20.7m/s = -9.19m/s

which means that now the ball is falling down.