Ask question

Ask question

All categories

3 years ago

12

A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building

under construction. A 75.0 kg construction worker stands at the far end of the beam. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. What is the magnitude of the torque about the point where the beam is bolted into place?

1 answer:

adell [148]3 years ago

6 0

Answer:

12164.4 Nm

Explanation:

CHECK THE ATTACHMENT

Given values are;

m1= 470 kg

x= 4m

m2= 75kg

Cm = center of mass

g= acceleration due to gravity= 9.82 m/s^2

The distance of centre of mass is x/2

Center of mass(1) = x/2

But x= 4 m

Then substitute, we have,

Center of mass(1) = 4/2 = 2m

We can find the total torque, through the summation of moments that comes from both the man and the beam.

τ = τ(1) + τ(2)

But

τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)

= 9221.4Nm

τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm

τ = τ(1) + τ(2)

= 9221.4Nm + 2943Nm

= 12164.4 Nm

Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm

You might be interested in

The standard wave format for any wave is wave. When depicting wave in standard wave format, the direction of motion must be rota

german

Answer:

Transverse wave and Longitudinal wave and Electromagnetic wave

Explanation:

An inverted wave is a wave in which the vibrations of the particles are perpendicular to the direction of wave motion.
Longitudinal waves, on the other hand, are waves in which the vibrations of the particles are parallel to the direction of wave motion.
Electromagnetic waves are waves that do not require medium media for transmission, including radio waves, microwaves, UV lights, etc.
Most electromagnetic waves are transverse in nature.

4 0

3 years ago

9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th

vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

$x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s$

Applying the kinematics equations for motion with uniform acceleration in x and y direction

So

$x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\ v_{o}Sin\alpha=21.828.....(2)$

Put the value of v₀ from equation (1) to equation (2)

So

$\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}$

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

$v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s$

7 0

3 years ago

Doss [256]

Answer:

The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.

Explanation:

Considering west direction along negative x-axis and north direction along positive y-axis

Given:

The car travels at a speed of 120 km/h in the west direction.

The car then travels at the same speed in the north direction.

Now, considering the given directions, the velocities are given as:

Velocity in west direction is, $\overrightarrow{v_1}=-120\ \vec{i}$

Velocity in north direction is, $\overrightarrow{v_2}=120\ \vec{j}$

Now, since $v_1\ and\ v_2$ are perpendicular to each other, their resultant magnitude is given as:

$|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}$

Plug in the given values and solve for the magnitude of the resultant.This gives,

$|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h$

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.

So, the direction is given as:

$x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)$

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.

4 0

3 years ago

A body 'A' of mass 1.5kg travelling along the positive X-axis with speed of 4.5m/s collides with another body 'B' of mass 3.2kg,

xz_007 [3.2K]

I already answered this question.
Please refer to this link brainly.com/question/8743596.

6 0

3 years ago

A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o

aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

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

Substitute,

14kg for m

9m/s for v

$E_r = \frac{1}{2} (14) (9)^2\\= 567J$

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

$E_R = \frac{1}{2} Iw^2$

The expression for the moment of inertia of the cylinder is,

$I = \frac{1}{2} mr^2$

The expression for angular velocity is,

$w = \frac{v}{r}$

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

$E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2$

$= \frac{mv^2}{4}$

$E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J$

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

$E = E_r + E_R\\\\$

substitute 567J for E(r) and 283.5J for E(R)

$E = 567J + 283.5\\= 850.5J$

The total energy of the cylinder is 850.5J

6 0

3 years ago

Read 2 more answers