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2 years ago

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

Physics

2 answers:

Keith_Richards [23]2 years ago

8 0

The hi.nge will apply a force of 142.10N on the beam in the vertical direction.

We must learn more about the tension in order to find the solution.

<h3>How can I determine the vertical component of the force the hi.nge has on the beam?</h3>

Let's create the system's free body diagram.

We must balance the total vertical force to zero in order to get the vertical component of the force applied to the beam by the height.

$F_V=mg-Tsin\alpha$

We must identify the tension in order to find the solution.

$Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N$

Consequently, the force that the height exerts on the beam will have a vertical component that is,

$F_v=(29*9.8)-(169.43*sin57)=142.10N$

This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.

Learn more about the tension here:

brainly.com/question/28106868

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castortr0y [4]2 years ago

5 0

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>

Let's draw the free body diagram of the system.
To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

$F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\$

To find the answer, we have to find the tension,

$Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N$

Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

$F_V=(29*9.8)-(169.43*sin57)=142.10N$

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

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Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

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2 years ago

Two trucks have the same mass. Which truck will have a greater amount of kinetic energy? b. The truck with less mass d. The truc

tatiyna

Answer:

c. The truck with greater speed

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents speed or velocity measured in metres per seconds square.

From the above formula, we can deduce that the kinetic energy of an object is directly proportional to the square of the value of its velocity or speed.

Hence, the truck with greater speed would have a greater amount of kinetic energy.

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2 years ago

I use 1 5/7 cans of paint to paint 7/8 of my room. How many cans of paint will I need to paint the entire room?

sergejj [24]

Answer:

y = 2.45 x

Explanation:

given,

Let 'x' be the number of cans.

and 'y' be the total area of the room

now,

From the given statement in the question

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$y = \dfrac{15}{7}\times \dfrac{8}{7} \times x$

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From the Above equation we can state that

To pain the entire room 2.45 cans of paint is required.

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3 years ago