Answer:
Explanation:
The elevator is moving with uniform speed and bolt will fall with acceleration due to gravity g .
Displacement of elevator when bolt starts falling
= 5.05 x 5.75 = 29 m
Let after time t bolt reaches the elevator
in time t , displacement of bolt = 1/2 g t²
= .5 x 9.8 x t² = 4.9 t²
displacement of elevator = 5.05 t
position of elevator = 29 + 5.05 t
According to question ,
29 + 5.05 t = 4.9 t²
4.9 t² - 5.05 t - 29 = 0
t = 3 s approx .
b )
Distance covered by the bolt
= 1 /2 g t²
= .5 x 9.8 x 9
= 44.1 m
Bolt must have to fall from a height of 44.1 m
or in terms of floor it is
44.1 / 3 = 15 th floor .
Convert the following to SI units: a) 75 in. b) 3.45 x 106 yr.c) 62 ft/day.d) 2.2 x 104 mi2.
Answer:
A. 0.025inches= 1m
So 75in= 0.025*75= 1.88m
B 3.45*10^6yrs= 3.45E6* 365x 86900s
= 1.09*10^8s
C. 62ft/day= 62* 0.3048/86900= 2.1*10^-4
D. 2.2*10^4mi² x ( 1609.3)²
= 6.1*10^7m²
A shopper walks eastward 3.2 meters and then westward
7.2 meters.
Pro
For this motion, what is the distance moved?
.A full length image of a distance tall building can definitely be seen by using
Answer:
a full length of image of a distance tall building can definitely be seen by using telescope
A sailboat took 25 hours to cover 1/4 of a journey. Then, it
covered the remaining 144 miles in 3.5 hours. What was the
average speed for the whole journey?
mph
Answer:
32
Explanation:
The average speed of the car for the remaining is 32 miles/h.
What is average speed?The average speed of any moving object is the ratio of the total distance covered and the total time taken to cover that distance.
Average Speed S= distance /time
Given is a sailboat took 2.5 hours to cover 1/4 of a journey. Then, it covered the remaining 144 miles in 3.5 hours.
If x be the total distance covered during journey, then
3/4 x = 144
x = 192 miles
The total time taken for journey is 2.5 +3.5 = 6 h
The average speed for the whole journey is
Avg speed S= 192 / 6
S = 32 miles/h
Thus, the average speed of the car for the remaining is 32 miles/h.
Learn more about average speed.
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Question 4 of 10
Which option is part of designing a set of experimental procedures?
O A. Developing a hypothesis
B. Determining how data will be gathered
C. Submitting conclusions to the peer-review process
O D. Beginning to take measurements
Answer:
Determining how data will be gathered
Explanation:
As a scientist considers the use of a particular set of experimental procedures to carry out his research, an important consideration is the method of data collection.
The experimental procedures chosen must be those that make the process of data collection much easier, efficient and lead to collection of reliable data.
Hence determining the method of data collection is very important when designing a set of experimental procedures.
A particle moves along the x-axis according to the equation
S = 4+6t-2t^2
where S is in meters and t is in seconds. At t = 3.0 s, calculate
(a) the position of the particle
(b) its instantaneous velocity
(c) its instantaneous acceleration.
◈ A particle is moving along the x-axis according to the equation
S = 4 + 6t - 2t²____________________________
✪ Position at t = 3s :
➝ S = 4 + 6t - 2t²
➝ S = 4 + 6(3) - 2(3)²
➝ S = 4 + 18 - 18
➝ S = 4m
✪ Instantaneous velocity at t = 3s :
➝ v = dx/dt = d(4 + 6t - 2t²)/dt
➝ v = 6 - 4t
➝ v = 6 - 4(3)
➝ v = 6 - [tex]\sf{12}[/tex]
➝ v = -12m/s
✪ Instantaneous acc. at t = 3s :
➝ a = dv/dt = d(6 - 4t)/dt
➝ a = -4m/s²
[Acceleration does not depend on time]
A plastic wire is wind on a wheel on a radius 3 cm. The whole system is initially at rest. A force is applied on the plastic wire and the wire unwinds from the wheel from its axis of rotation. The wheel is given an angular acceleration of 92.00 rad/s2 for:______.
Answer:
The torque of the wheel is 0.92 N-m.
Explanation:
Given that,
Radius of wire = 3 cm
Angular acceleration = 92.00 rad/s²
Suppose, Mass of wheel is 4 kg and radius is 5 cm. find the torque of the wheel.
We need to calculate the torque of the wheel
Using formula of torque
[tex]\tau=I\alpha[/tex]
[tex]\tau=MR^2\times\alpha[/tex]
Where, M = mass of wheel
R = radius of wheel
[tex]\alpha[/tex] = angular acceleration
Put the value into the formula
[tex]\tau=4\times(5\times10^{-2})^2\times92[/tex]
[tex]\tau=0.92\ N-m[/tex]
Hence, The torque of the wheel is 0.92 N-m.
Sympathy, Unproductive, Perspectives
Empathetic people try to look at things from different _______
People with low empathy often are unable to understand others perspective and involve in _____arguments.
_________ is when you feel sorry for the other person and there is no urge, no impelling reason to be feel others feelings.
Answer:
Empathetic people try to look at things from different perspectives
People with low empathy often are unable to understand others perspective and involve in unproductive arguments.
sympathy is when you feel sorry for the other person and there is no urge, no impelling reason to be feel others feelings.
Explanation:
A car is moving at 6 m/s and then accelerates at 1.7 m/s2 for 4.2 seconds. What is the final velocity of the car?
Explanation:
Hey there!!
Here,
Initial velocity (u) = 6m/s.
Acceleration (a) = 1.7m/s^2.
Time (t) = 4.2s.
final velocity (v) = ?
We have,
[tex]a = \frac{v - u}{t} [/tex]
Putting their values,
[tex]1.7 = \frac{v - 6}{4.2} [/tex]
7.14 = v - 6
v = 7.14 + 6
Therefore, the final velocity is 13.14 m/s.
Hope it helps....
i NED HELP i’m so confused
A machine, modeled as a simple spring-mass system, oscillates in simple harmonic motion. Its acceleration is measured to have an amplitude of 10,000 mm/s2 with a frequency of 8 Hz. Compute the maximum displacement the machine undergoes during this oscillation.
Answer:
3.96*10^-3 m
Explanation:
Given that
Amplitude, A = 10000 mm/s² = 10 m/s²
Frequency, f = 8 hz
the maximum acceleration of a simple harmonic motion is given as
|a|(max) = Aw², where
A = amplitude of the simple harmonic motion
w = angular frequency of the simple harmonic motion
Remember, w = 2πf, so
w = 2 * 3.142 * 8
w = 50.272 rad/s
Rearranging the formula, and making a the subject, we have
A = |a|(max) / w², so that
A = 10 / 50.272²
A = 10 / 2527.27
A = 0.00396 m
A = 3.96*10^-3 m
Which word describes a statement that has been accepted, tested, and supported by multiple sets of evidence? A. Hypothesis B. Theory C. Procedure D. Data
Answer:
Data
Explanation:
can someone define Newton's 1st 2nd and 3rd law
Answer:
HEY THERE! HERE'S YOUR ANSWER!
Explanation:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
HOPE THIS HELPS YOU BUDDY!
A train is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. Calculate the horizontal velocity of the train at the end of 3 seconds.
Answer:
57,9 m/s
Explanation:
A train is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. The horizontal velocity of the train at the end of 3 seconds is 100m/s.
The first formula we will use is:
v = d/t
300m ÷ 5s = 60m/s
The last formula we will use is:
vh = v·cosx
60m/s·cos15.0°
60m/s · 0,965 = 57,9 m/s
Therefore, the answer is 57,9 m/s.
Hope this helped!
57,9 m/s
Explanation:velocity formula
v = d/t
= 300m/5s
= 60m/s
horizontal velocity formula
vh = v·cosx
= 60m/s×cos15°
= 60m/s×0,965
= 57,9 m/s
A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t2 + 8 underroot t, where t is the time in seconds and s(t) is the distance in feet from the top of the building. A) Write a formula for the average velocity of the ball for t near 4. B) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds. C) What is your estimate for the instantaneous velocity of the ball at t = 4.
Answer:
Explanation:
s(t) = -16t2 + 8√t
A )
Average velocity
s(t) / t = (-16t2 + 8√t)/t
A(t)= -16t + 8 / √t
average velocity of the ball for t near 4.
A(t) = -16t + 8 / √t
Lt t⇒4
B )
Distance covered in 4 s
-16t2 + 8√t
= - 16 x 16 + 8 x 2
= - 240
Distance covered in 5 s
= - 16 x 25 + 8 √5
= -400 + 17.88
= -382.12
distance covered in duration from 4 to 5 sec
= -142.12
velocity = - 142.12 / 1 = - 142.12 m /s
Distance covered in 4.5 s
= -16 x 4.5² + 8√4.5
= -324 + 16.97
= -307
distance covered during 4 to 4.5
= 67
velocity during 4 to 4.5
= -67 / .5
- 134 m /s
distance covered in 4.05 s
-16 x 4.05² + 8√4.05
-262.44 + 16.1
-246.34
distance covered during 4 to 4.05
= -6.34
velocity during 4 to 4.05
= -6.34 / .05
- 126.8 m /s
C )
Instantaneous velocity at t = 4
= - 120 m /s
(A) As 't' approaches to 4s, the formula of average velocity is
[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]
(B) The average velocity for the time interval starting at t = 4 with a duration of 1 second is [tex]142.11\,m/s[/tex].
The average velocity for the time interval starting at t = 4 with a duration of 0.5 seconds is [tex]-134.058\,m/s[/tex]
The average velocity for the time interval starting at t = 4 with a duration of 0.05 seconds is [tex]-126.8\,m/s[/tex].
(C) The instantaneous velocity of the ball at 4s is [tex]-126\,m/s[/tex].
The answers are explained as follows.
Given that the distance is a function of time.
[tex]s(t)=-16\,t^2\,+\,8\sqrt{t}[/tex]
(A) The average velocity can be given by,
[tex]v_{avg}=\frac{-16\,t^2\,+\,8\sqrt{t} }{t} =-16t+ \frac{8}{\sqrt{t}}[/tex]
As 't' approaches to 4s, the formula becomes;
[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]
(B) We know that the average velocity or in this case speed is the total distance by the total time taken.
Distance covered in 4 s can be found by putting [tex]t=4s[/tex] in the distance formula.
[tex]s(4)=(-16\times16)+(8\times2)=-240\,m[/tex]Distance covered in 5 s can also be found by the same method
[tex]s(5)=(-16\times25)+(8\times\sqrt{5} )=-382.11\,m[/tex]Therefore, the distance covered in from 4 to 5 seconds is;
[tex]s(5) -s(4)=-382.11\,m-(-240\,m)=-142.11\,m[/tex]So, the average velocity here = [tex]\frac{-142.11\,m}{5\,s-4\,s}=-142.11\,m/s[/tex]Distance covered in 4.5 s is given by,
[tex]s(4.5)=(-16\times4.5^2)+(8\times\sqrt{4.5} )=-307.029\,m[/tex]Therefore, the distance covered in 4 to 4.5 seconds is;
[tex]s(4.5)-s(4)=-307.029\,m-(-240\,m)=67.029\,m[/tex]So, the average velocity here = [tex]\frac{-67.029\,m}{4.5\,s-4\,s}=-134.058\,m/s[/tex]Distance covered in 4.05 s is given by,
[tex]s(4.05)=(-16\times4.05^2)+(8\times\sqrt{4.05} )=-246.34\,m[/tex]Therefore, the distance covered in 4 to 4.05 seconds is;
[tex]s(4.05)-s(4)=-246.34\,m-(-240\,m)=-6.34\,m[/tex]So, the average velocity here = [tex]\frac{-6.34\,m}{4.05\,s-4\,s}=-126.8\,m/s[/tex](C) The instantaneous velocity of the ball can be found by differentiating the function [tex]s(t)[/tex].
[tex]v(t)=\frac{ds(t)}{dt} =\frac{d}{dt}(-16\,t^2\,+\,8\sqrt{t})=-32t+\frac{4}{\sqrt{t} }[/tex]For the instantaneous velocity of the ball at 4s, substitute [tex]t=4\,s[/tex] in the above equation.[tex]v(t)=(-32\times4)+\frac{4}{2}=-126\,m/s[/tex]Learn more about finding the velocity at a given time here:
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A flea jumps straight up to a maximum height of 0.500 m . What is its initial velocity v0 as it leaves the ground? Express your answer in meters per second to three significant figures.
Answer:
v0 = 3.13m/sExplanation:
Using the equation of motion to get the initial velocity as it leaves the ground. According to the equation of motion;
v² = v0²+2as where;
v is the final velocity
v0 is the initial velocity
a is the acceleration due to gravity
s is the height.
At the maximum height, the final velocity v = 0m/s
Given s = 0.500m and a = g = -9.81m/s² (acceleration due to gravity is negative due t the upward movement of the flea)
0² = v0² - 2(9.81)(0.500)
0 = v0² - 9.81
- v0² = -9.81
v0 = √9.81
v0 = 3.13m/s
Hence the initial velocity v0 of the flea as it leaves the ground is 3.13m/s (to three significant figures)
"Find the total weight of an 18-ft3 tank of oxygen if the oxygen is pressurized to 184 psia, the tank itself weighs 150 lbf, and the temperature is 95°F."
Answer:
Explanation:
1 psi = 6894.76 Pa
P = 184 psi = 12.686 x 10⁵ Pa .
Temperature T = 95⁰F = 35⁰C= 308 K
volume V = 18 ft³ = 18 x 0.0283168 m³
= .51 m³
From the gas law
PV/RT = n where n is mole of gas
= 12.686 x 10⁵ x .51 / 8.31 x 308
= 252.78 gm mole
= 252.78 x 32 gm
= 8.08896 kg
= 2.20462 x 8.08896 lb
= 17.833 lb
= 17.833 / 32 lbf
= .5573 lbf
weight of tank = 150 lbf
Total weight = 150 + .5573
= 150.5573 lbf .
A soccer ball is at rest. A player kicks the ball from the ground. What impact will the energy transfer have on the ball?
A. The ball will stay at rest
B. The ball will move forward
C. The ball will move backward.
D. The ball will stay in the same position.
Answer:
Its B
Explanation:
the force of the ball is being kicked from the back side of the ball will causes a pushing motion and the ball will move forward
A softball player is testing her physics knowledge and tosses a softball upward from the top of a building. The building is 100 m tall and the ball starts with a velocity of 10 m/s.a. What is the maximum height the ball reaches and at what time? b. How much time does it take for the ball to go from h=50 m to h=0 m on the way down? c. What is the velocity when h=50 m?
Answer:
Explanation:
We shall apply newton's laws formula
a )
initial velocity in upward direction u = 10 m/s
acceleration due to gravity g = 9.8/ m .s²
Let h be the maximum height where v = o
v² = u² - 2gh
0 = 10² - 2 gh
h = 10² / 2g
= 10² / 2 x 9.8
= 5.10 m
Since the ball was thrown from height of 100 m , total maximum height of ball
= 100 + 5.10
= 105.10 m
Let t be the time taken
v = u - gt
0 = 10 - gt
t = 10 / 9.8
= 1.02 s
b )
when h = 50 on its way downwards , velocity
v² = u² + 2 g s
v² = 0 + 2 x 9.8 x ( 105.10 - 50 )
[ distance travelled by ball at this point from top = 105.1 - 50 = 55.10 ]
v = 32.86 m / s
Let us find out final velocity of touching the ground . For it distance travelled = 105.10
v² = u² + 2gh
v² = 0 + 2 x 9.8 x 105.1
v = 45.39 m /s
Now velocity at h = 50 is 32.86
velocity at h = 0 is 45.39
time taken to travel fro h = 50 to h = 0
v = u + gt
45 .39 = 32.86 + 9.8 x t
t = 1.28 s .
The element found in Period 2, Group 8 is called Argon. True False
Tony completed a 720 km journey with an average speed of
90 km/h. For the first 285 km, he travelled at an average speed
of 95 km/h. What was the average speed for the remaining
journey?
Answer:
87 km/h
Explanation:
a graph compares the weight and height of several different breeds of cat. What type of model is the graph?
Answer: Mathematical Model
Explanation:
Took the test
What is true about all uranium atoms
Answer:
erm they all should have same numbers of protons
Answer:
They each have the same number of nuclear particles.
Explanation:
What is true about all uranium atoms? They each have the same number of nuclear particles. They each have the same number of neutral particles. They each have the same number of neutrons . Uranium is an element that is often used in nuclear power plants. I hope this helped :)
What do you think is the
difference between
speed and acceleration?
What do you think is the
difference between mass
and weight?
a quarterback throws a football in a high, arching spiral to a receiver running down the field. where is the football traveling the fastest? where is it traveling the slowest?
Answer:
Travelling fastest: when the football leaves the hands of the quarterback and when the receiver gets it
Travelling slowest: at the very top of then arched trajectory
Explanation:
Notice that this example can be analyzed with a two dimensional pattern as in the launching of a cannonball, where we study separately the velocities and acceleration acting vertically and horizontally.
The football is thrown at an agle with the horizontal with an initial velocity that is decomposed in the vertical and in the horizontal axes.
The Horizontal movement of the football is that of an object with constant velocity (that of the horizontal component of the initial velocity imparted by the quarterback) .
The vertical movement of the football is that of an object moving in an accelerated fashion, with constant acceleration due to the gravitational field, and with an initial velocity opposite to that constant acceleration. The initial velocity is that of the vertical component of the initial velocity imparted by the quarterback.
AT every point on the path of the ball, its velocity can be calculated by the vector addition of the horizontal component (with constant velocity as we discussed above) and the vertical component (this velocity is changing since its is under accelerated motion) of the object's velocity at any given time through the path.
The maximum velocity of the football will be at the point where the two components are at their maximum (that is when the football leaves the hand of the thrower, and when it gets to the hands of the receiver.
While the minimum is going to be when the vertical component of the velocity is at is minimum (zero) at the top of the arched trajectory when the football changes vertical direction and starts heading downwards.
4. Describe how the velocity of an object changes if it undergoes uniformly acceleration motion. Can its direction change?
Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
[tex]a_{t}[/tex] = (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
[tex]a_{c}[/tex] = v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)
The velocity of an object changes if it undergoes uniformly acceleration
motion by considering if it is:
Positive constant accelerationNegative constant accelerationWhat is Velocity?This is the rate of change of an object's position with time. If the object has
positive constant acceleration, the slope goes upward while if it is a
negative constant acceleration, the slope goes downward.
The direction doesn't change as a result of the uniform speed and
direction that is involved.
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Daniel has a bill of 2750 on his credit card. If interest is charged at a rate of 15% p.a., calculate the amount of interest that Daniel must pay for the month?
Answer:
412.5
Explanation:
all you have to do is multiply .15
what is a hypothisis
Answer:
a precise, testable statement of what the reseatcher predict will be the outcome of the study
Explanation:
Answer:
supposition or proposed explanation
Explanation:
its made on the basis of limited evidence as a starting point for further investigation.
If a person has a velocity of 5 meters per second, how much time will they take to travel 1.5 kilometers
Answer:
300 secsExplanation:
Velocity is the change in displacement of a body with respect to time.
Velocity = Displacement/Time
Given velocity = 5m/s
Displacement = 1.5 km
Displacement = 1.5km * 1000 = 1500m
From the formula, Time = Displacement/Velocity
Time = 1500/5
Time = 300 secs
Hence the time it will take to travel 1.5km is 300secs
1)Determine, in terms of unit vectors, the resultant of the five forces illustrated in the figure, Consider F1=20 N, F2= 12 N, F3=17 N. F4=15N and F5=9N
2)In the situation illustrated below, two blocks of Ma and Mb masses are suspended between two walls using ropes. If Ma=2kg, calculate the tension developed in the segments AB, BC and CD and the mass of block B for the equilibrium condition
3)In the figure below, the crane and its AB jib have together 390kg and center of mass in G1. If the 90kg BCD cage and the 80kg man have center of mass located in G2 and G3, respectively, (a) calculate the angle θ of inclination of the boom for which the crane is on the eminence of tumbling. (b) for which size the boom should be reduced (retracting its point B) so that it can be positioned horizontally (θ=0) and the crane does not tip over. Consider that the cage always stays in the horizontal position
4) A 1.7kg plate with center of mass in G is supported by a bar and three cables, as shown in the figure below. Determine the voltage developed in the AB, AC and DE cables and the reaction (there is only one) at point O for the equilibrium condition
Explanation:
1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis. The angle between F₁ and the +z axis is 30°. Therefore, the vector is:
<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)
<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)
<F₁> = 5 i + 5√3 j + 10√3 k
F₂ is in the xy plane. Its slope is -24/7. The vector is:
<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)
<F₂> = -3.36 i + 11.52 j
F₃ is parallel to the +x axis. The vector is:
<F₃> = 17 (i + 0 j + 0 k)
<F₃> = 17 i
F₄ is parallel to the -z axis. The vector is:
<F₄> = 15 (0 i + 0 j − k)
<F₄> = -15 k
F₅ is in the xy plane. It forms a 15° angle with the -y axis. The vector is:
<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)
<F₅> = -9 sin 15° i − 9 cos 15° j
The resultant vector is therefore:
<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k
<F> = 16.31 i + 11.49 j + 2.32 k
2) Sum of forces at point B in the x direction:
∑F = ma
Tbc cos 40° − ¹⁵/₁₇ Tab = 0
Tbc cos 40° = ¹⁵/₁₇ Tab
Tbc = 1.15 Tab
Sum of forces at point B in the y direction:
∑F = ma
Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0
Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)
(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N
1.21 Tab = 20 N
Tab = 16.52 N
Tbc = 19.02 N
Sum of forces at point C in the x direction:
∑F = ma
Tcd sin 25° − Tbc cos 40° = 0
Tcd sin 25° = Tbc cos 40°
Tcd = 1.81 Tbc
Tcd = 34.48 N
3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F. Relative to point A:
3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)
2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm
1336 kgm = 1530 kgm cos θ
θ = 29.17°
3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)
2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm
1336 kgm = (170 kg) x
x = 7.86 m
4) Find the lengths of the cables.
Lab = √((2 m)² + (3 m)² + (5 m)²)
Lab = √38 m
Lac = √((2 m)² + (3 m)² + (5 m)²)
Lac = √38 m
Lde = √((2 m)² + (3 m)²)
Lde = √13 m
Sum of forces in the x direction:
∑F = ma
-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0
Sum of forces in the y direction:
∑F = ma
2/√38 Fab − 2/√38 Fac = 0
Fab = Fac
Sum of forces in the z direction:
∑F = ma
3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0
Sum of moments about the y-axis:
∑τ = Iα
(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0
Substitute Fab = Fac and simplify:
6/√38 Fab + 3/√13 Fde − mg = 0
30/√38 Fab + 6/√13 Fde − 2mg = 0
Double first equation:
12/√38 Fab + 6/√13 Fde − 2mg = 0
Subtract from the second equation:
28/√38 Fab = 0
Fab = 0
Fac = 0
Solve for Fde:
3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0
3/√13 Fde = mg
3/√13 Fde = (1.7 kg) (10 m/s²)
Fde = 20.43 N
Solve for Rx:
-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0
Rx = 2/√13 Fde
Rx = 11.33 N