Ask question

Ask question

All categories

4 years ago

5

A bar extends perpendicularly from a vertical wall. The length of the bar is 2 m, and its mass is 10 kg. The free end of the rod

is attached to a point on the wall by a light cable, which makes an angle of 30° with the bar. Find the tension in the cable.

1 answer:

Simora [160]4 years ago

6 0

Answer:

T = 98.1 N

Explanation:

Given:

- mass of bar m = 10 kg

- Length of the bar L = 2 m

- Angle between Cable and wall Q = 30 degres

Find:

- Find the tension in the cable.

Solution:

- Take moments about intersection of bar and wall, to be zero (static equilibrium)

(M)_a = 0

T*sin( 30 )*L - m*g*L/2 = 0

T*sin(30) - m*g / 2 = 0

T = m*g / 2*sin(30)

T = 10*9.81 / 2*sin(30)

T = 98.1 N

You might be interested in

Estimate the kinetic energy of the earth with respect to the sun as the sum of two terms.

nekit [7.7K]

The definition of kinetic energy allows to find the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is $\frac{K_{Sum} }{K_{Earth}} = 5.3 \ 10^2$

<h3 /><h3 /><h3> Kinetic enrgy.</h3>

Kinetic energy is the energy due to the movement of bodies, it is given by the relation

K = ½ m v²

where K is the kinetic energy, m the mass of the body and v the velocity of the body.

In a compound motion it is common to separate energy into parts to simplify calculations.

Translational kinetic energy. Due to the linear movement of the body

$K_{tras} =\frac{1}{2} m v^2$

Rotational kinetic energy. Due to the rotational movement of the body.

$K_{rot} = \frac{1}{2} I w^2$

Where I is the inrtia momentum and w the angular velocity.

They indicate that we compare the kinetic energy of the sun and the Earth.

The Earth has two movements, one of rotation about its axis with a period of T = 24 h and one of translation with respect to the Sun with a period of T= 365 days, therefore the kinetic energy of the Earth.

$K_{earth} = K_{tras} + K_{rot}$

Linear and rotational speed are related.

v = w r

The Earth is an almost spherical body therefore the moment of inertia of a solid sphere.

I = $\frac{2}{5 } m r^2$

Let's subatitute.

$K_{earth} = \frac{1}{2} \ m r^2_{tras} w^2_{tras} + \frac{1}{2} ( \frac{2}{5} m r^2_{earth}) w^2_{rot}$

The movement of the Earth around the sun is almost circular, therefore we can use the relations of the uniform circular movement, where the angle for one revolution is 2π radians and the time is called the period.

$w = \frac{2 \pi}{T}$

Let's substitute.

$K_{earth} = \frac{1}{2} m ( \frac{2\pi r^2_{tras}}{T_{tras}})^2 \ + \frac{1}{5} m (\frac{2\pi r^2_{earth} }{T^2_{rot}})^2$

$K_{earth} = 4 \pi^2 \ m \ ( \frac{1}{2} [ \frac{r_{tras}}{T_{tras}y} ]^2 + \frac{1}{5} [ \frac{r_{rot}}{T_{rot}}]^2)$

Data for Earth are tabulated:

Mass m = 5.98 1024 kg

Radius r = 6.37 10⁶ m

Radius orbits tras = 1.496 10¹¹ m

Rotation period $T_{rot}$ = 24 h ( $\frac{3600s}{1h}$ ) = 8.64 10⁴s

Translation period $T_{tras}$ = 365 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 3.15 10⁷ s

Let's calculate.

$K_{earth} = 4 \pi^2 5.98 \ 10^{24} ( \frac{1}{2} ( \frac{1.496 \ 10^{11}}{3.15 \ 10^7 } )^2 \ + \frac{1}{5}( \frac{6.37 \ 10^6 }{8.64 \ 10^4})^2 )$

$K_{earth} = 2.36 \ 10^{26 } \ (1.128 \ 10^7 + 1.087 \ 10^3)$

$K_{earth}= 2.66 \ 10^{33} J$

Let's analyze the kinetic energy for the Sun, this is inside the solar system therefore it has no translation movement and is approximately a sphere with a rotation period of $T_{Sum}$ = 27 days.

The kinetic energy of the sun is;

$K_{sum} = K_{rot} = \frac{1}{2} I w^2$

$K_{sum} = \frac{1}{2} (\frac{2}{5} M R^2) (\frac{2\pi}{T_{sum}})^2$

$K_{sum} = \frac{4\pi^2 }{5} M (\frac{R}{T_{rot}})^2$

The tabulated data for the sun are:

Mass m = 1,991 1030 kg.

Radius R = 6.96 10⁸ m

Period T = 27 d ( $\frac{24h}{1 d}$ ) ( $\frac{3600s}{1h}$ ) = 2.33 10⁶ s

Let's calculate.

$K_{sum} = 1.40 \ 10^{36} J$

The relationship of the kinetic energy of the sun and the Earth is:

$\frac{K_{sum}}{K_{earth}} = \frac{1.40 \ 10^{36}}{2.66 \ 10^{33}}$

$\frac{K_{sum}}{K_{earth}} = 5.3 \ 10^2$

In conclusion using the definition of kinetic energy we can shorten the result for the relationship between the energy of the sun and the Earth is:

The kinetic energy ratio is: $\frac{K_{Sum}}{K_{Earth}} = 5 \ 10^2$

Learn more about kinetic energy here: brainly.com/question/25959744

5 0

3 years ago

Careers in this career cluster also include related professional and technical support activities such as production planning an

qwelly [4]

True, all of those careers go in the same category

8 0

4 years ago

Características sobre los conectores SATA

Soloha48 [4]

Answer:

SATA tiene las siguientes características importantes: Voltaje reducido. Los estándares ATA actuales utilizan 5,0 V o 3,3 V (ATA-100/133). Estos voltajes relativamente altos junto con altas densidades de pines hacen que 100 MB / s sea la velocidad de datos más alta que se puede alcanzar de manera realista.

6 0

3 years ago

Charcoal appears black in color because it

steposvetlana [31]

C. This is the idea of "black body radiation". Charcoal/carbon is a "perfect black body". Absorbs all radiation. There's a whole host of stuff about this in physics, including, I think, Planck's (Nobel Prize winner) black body radiation theory.

4 0

3 years ago

Read 2 more answers

When a substance goes directly from a gaseous state to a solid state as dry ice does

Naddik [55]

Answer: Sublimation I think.

Explanation:

6 0

3 years ago

Read 2 more answers