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Pachacha [2.7K]

3 years ago

11

A strong weightless rope has a mass, m, hanging from the middle of it. The tension force on each rope is 25 N, and the rope droo

ps at an angle of 20.0 degrees. How much mass is hanging from the rope?

1 answer:

bezimeni [28]3 years ago

4 0

By using Lami's theorem, Mass m = 1.75 kg approximately

Given that a strong weightless rope has a mass, m, hanging from the middle of it. If the tension force on each rope is 25 N, and the rope droops at an angle of 20.0 degrees to the horizontal.

By using Lami's theorem, we can get how much mass is hanging from the rope.

Let the angle between the rope = α = 180 - 40

α = 140 degrees

The angle between one of the rope and mass = β = 20 + 90

β = 110 degrees

The angle between the mass and the other rope = γ = 360 - (140 + 110)

γ = 360 - 250

γ = 110 degrees

W/ sinα = T/ sinβ = T/sinγ

W/ sinα = T/ sinβ

Substitute all the necessary parameters

W/sin140 = 25/sin 110

W / 0.643 = 25 / 0.939

W = 17.1 N

Weight W = mg

17.1 = 9.8m

mass m = 17.1/9.8

Mass m = 1.7455 kg

Mass m = 1.75 kg approximately

Therefore, 1.75 kg mass is hanging from the rope.

Learn more about resolution of forces here: brainly.com/question/1858958

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A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

a)

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

b)

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

c)

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

d)

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

e)

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

f)

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago