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3 years ago

Rigid Body Statics in 3 Dimensions

Physics

1 answer:

slamgirl [31]3 years ago

7 0

Explanation:

Draw a free body diagram of the bar.

There are 3 reaction forces at O in the x, y, and z direction (Ox, Oy, and Oz).

There is a tension force Tac at A in the direction of the rope. There are also tension forces Tbd and Tbe at B in the direction of the ropes.

Finally, there is a weight force mg pulling down halfway between A and B, where m = 400 kg.

There are 6 unknown variables, so we'll need 6 equations to solve. Summing the forces in the x, y, and z direction will give us 3 equations. Summing the torques about the x, y, and z axes will give us 3 more equations.

First, let's find the components of the tension forces.

Tbe is purely in the z direction.

Tbd has components in the y and z directions. The length of Tbd is √8.

(Tbd)y = 2/√8 Tbd

(Tbd)z = 2/√8 Tbd

Tac has components in the x, y, and z directions. The length of Tac is √6.

(Tac)x = 1/√6 Tac

(Tac)y = 1/√6 Tac

(Tac)z = 2/√6 Tac

Sum of the forces in the +x direction:

∑F = ma

Ox − (Tac)x = 0

Ox − 1/√6 Tac = 0

Sum of the forces in the +y direction:

∑F = ma

Oy + (Tac)y + (Tbd)y − mg = 0

Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0

Sum of the forces in the +z direction:

∑F = ma

Oz − (Tac)z − (Tbd)z − Tbe = 0

Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0

Sum of the torques counterclockwise about the x-axis:

∑τ = Iα

mg (2 m) − (Tac)y (2 m) − (Tbd)y (2 m) = 0

mg − (Tac)y − (Tbd)y = 0

mg − 1/√6 Tac − 2/√8 Tbd = 0

Sum of the torques counterclockwise about the y-axis:

∑τ = Iα

-(Tac)x (2 m) + (Tbd)z (1.5 m) + Tbe (1.5 m) = 0

-4 (Tac)x + 3 (Tbd)z + 3 Tbe = 0

-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0

Sum of the torques counterclockwise about the z-axis:

∑τ = Iα

-mg (0.75 m) + (Tbd)y (1.5 m) = 0

-mg + 2 (Tbd)y = 0

-mg + 4/√8 Tbd = 0

As you can see, by summing the torques about axes passing through O, we were able to write 3 equations independent of those reaction forces. We can solve these equations for the tension forces, then go back and find the reaction forces.

-mg + 4/√8 Tbd = 0

4/√8 Tbd = mg

Tbd = √8 mg / 4

Tbd = √8 (400 kg) (9.8 m/s²) / 4

Tbd = 2772 N

mg − 1/√6 Tac − 2/√8 Tbd = 0

1/√6 Tac = mg − 2/√8 Tbd

Tac = √6 (mg − 2/√8 Tbd)

Tac = √6 ((400 kg) (9.8 m/s²) − 2/√8 (2772 N))

Tac = 4801 N

-4/√6 Tac + 6/√8 Tbd + 3 Tbe = 0

3 Tbe = 4/√6 Tac − 6/√8 Tbd

Tbe = (4/√6 Tac − 6/√8 Tbd) / 3

Tbe = (4/√6 (4801 N) − 6/√8 (2772 N)) / 3

Tbe = 653 N

Now, using our sum of forces equations to find the reactions:

Ox − 1/√6 Tac = 0

Ox = 1/√6 Tac

Ox = 1/√6 (4801 N)

Ox = 1960 N

Oy + 1/√6 Tac + 2/√8 Tbd − mg = 0

Oy = mg − 1/√6 Tac − 2/√8 Tbd

Oy = (400 kg) (9.8 m/s²) − 1/√6 (4801 N) − 2/√8 (2772 N)

Oy = 0 N

Oz − 2/√6 Tac − 2/√8 Tbd − Tbe = 0

Oz = 2/√6 Tac + 2/√8 Tbd + Tbe

Oz = 2/√6 (4801 N) + 2/√8 (2772 N) + 653 N

Oz = 6533 N

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When balanced forces act on an object, what type of motion can the object have?

Lesechka [4]

Answer: A. It can only be motionless.

Explanation: When balanced forces act on an object, the object will not move. The only way that object can move is if the forces become unbalanced. Suppose there is a shoe on a table. You are pulling back on the shoe with a force of 3 newtons. But the friction from the shoe rubbing on the table is also 3 newtons of force. But the gravity pulling down on the shoe is also 3 newtons. Nothing will move. The only way the shoe can move is if you pulled with a little more force. The shoe would then move because the force you are pulling with is greater than the other forces acting on the shoe.

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3 years ago

There are many well-documented cases of people surviving falls from heights greater than 20.0 m. In one such case, a 55.0 kg wom

bixtya [17]

1a) -192.7g

1b) 0.0126 s

2) 1309 kg m/s

3) $1.04\cdot 10^5 N$

Explanation:

1a)

First of all, we have to find the velocity of the womena just before hitting the ground.

Since the total mechanical energy is conserved during the fall, the initial gravitational potential energy of the woman when she is at the top is entirely converted into kinetic energy.

So we can write:

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

where

m = 55.0 kg is the mass of the woman

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

h = 29.0 m is the initial height of the woman

v is her final speed

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(29.0)}=23.8 m/s$

Then, when the woman hits the soil, she is decelerated until a final velocity

$v'=0$

So we can find the deceleration using the suvat equation:

$v'^2-v^2=2as$

where

s = 15.0 cm = 0.15 m is the displacement during the deceleration

Solving for a,

$a=\frac{v'^2-v^2}{2s}=\frac{0-23.8^2}{2(0.15)}=-1888.3 m/s^2$

In terms of g,

$a=\frac{-1888.3}{9.8}=-192.7g$

1b)

Here we want to find the time it takes for the woman to stop.

Since her motion is a uniformly accelerated motion, we can do it by using the following suvat equation:

$v'=v+at$

where here we have:

v' = 0 is the final velocity of the woman

v = 23.8 m/s is her initial velocity before the impact

$a=-1888.3 m/s^2$ is the acceleration of the woman

t is the time of the impact

Solving for t, we find:

$t=\frac{v'-v}{a}=\frac{0-23.8}{-1888.3}=0.0126 s$

So, the woman took 0.0126 s to stop.

The impulse exerted on an object is equal to the change in momentum experienced by the object.

Therefore, it is given by:

$I=\Delta p =m(v'-v)$

where

$\Delta p$ is the change in momentum

m is the mass of the object

v is the initial velocity

v' is the final velocity

Here we have:

m = 55.0 kg is the mass of the woman

v = 23.8 m/s is her initial velocity before the impact

v' = 0 is her final velocity

So, the impulse is:

$I=(55.0)(0-23.8)=-1309 kg m/s$

where the negative sign indicates the direction opposite to the motion; so the magnitude is 1309 kg m/s.

The impulse exerted on an object is related to the force applied on the object by the equation

$I=F\Delta t$

where

I is the impulse

F is the average force on the object

$\Delta t$ is the time of the collision

Here we have:

$I=1309 kg m/s$ is the magnitude of the impulse

$\Delta t = 0.0126 s$ is the duration of the collision

Solving for F, we find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{1309}{0.0126}=1.04\cdot 10^5 N$

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2. Another car weighs 2000kg, you can push it .05 m/s?, how much force are you

Vesna [10]

Answer:

$\boxed {\boxed {\sf 100 \ Newtons}}$

Explanation:

We are asked to calculate the force you are applying to a car. According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, we can use the following formula to calculate force.

$F= m \times a$

The mass of the car is 2000 kilograms and the acceleration is 0.5 meters per second squared.

m= 2000 kg
a= 0.05 m/s²

Substitute the values into the formula.

$F= 2000 \ kg \times 0.05 \ m/s^2$

Multiply.

$F= 100 \ kg *m/s^2$

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 100 kilogram meters per second square is equal to 100 Newtons.

$F= 100 \ N$

You apply <u>100 Newtons</u> of force to the car.

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3 years ago

An experiment in which alpha particles were used to bombard thin sheets of gold foil led to the conclusion that an atom is compo

fiasKO [112]

Answer:

a small dense nucleus with positive charge and the nucleus is surrounding by orbiting electrons.

Explanation:

An experiment in which alpha particles were used to bombard thin sheets of gold foil led to the conclusion that an atom is composed mostly of a small dense nucleus with positive charge and the nucleus is surrounding by orbiting electrons.

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