Answer:
c)Average acceleration is always equal to the change in velocity of an object divided by the time interval.
Explanation:
Because average acceleration= a2+a1/2
Which is equal to change in velocity which is a time rate of change of velocity v-u/t which actually explains average acceleration
What do you think is the
difference between
speed and acceleration?
What do you think is the
difference between mass
and weight?
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 .
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....
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 cylindrical flask is fitted with an airtight piston that is free to slide up and down. A mass rests on the top of the piston. The initial temperature of the system is 313 K and the pressure of the gas is held constant at 137 kPa. The temperature is now increased until the height of the piston rises from 23.4 cm to 27.0 cm.What is the final temperature of the gas?
Answer:
T2 = 361.15 K
Explanation:
Since pressure is held constant, then we'll use Charles law
Charles law gives that V1/T1 = V2/T2
We are given;
Initial temperature; T1 = 313 K
Initial height; h1 = 23.4 cm
Final height; h2 = 27.0 cm
Now, formula for volume of a cylinder is V = πr²h
V1 = πr²h1 and V2 = πr²h2
Now, since V1/T1 = V2/T2 and we are looking for T2, let's make T2 subject;
T2 = (V2 × T1)/V1
Plugging in the relevant values gives;
T2 = (πr²h2 × 313)/πr²h1
πr² will cancel out to give;
T2 = (h2 × 313)/h1
Putting 23.4 for h1 and 27 for h2 gives;
T2 = (27 × 313)/23.4
T2 = 361.15 K
all digits shown on the measuring device, plus one estimated digit. are consided what
Answer:
The significant figures in a measurement consist of all the certain digits in that measurement plus one uncertain or estimated digit. In the ruler illustration below, the bottom ruler gave a length with 2 significant figures, while the top ruler gave a length with 3 significant figures.
Explanation:
3)What are the "bright stars" that seem to move against the sky? Are they really stars? What are they?
.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
The element found in Period 2, Group 8 is called Argon. True False
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 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.
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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PLEASE HELP! BRAINLIEST! The three main areas of science each contribute to forensic science in unique ways. Using your own words, write a paragraph (five or more sentences) to describe how biological sciences, chemical sciences, and physical sciences can contribute to the analysis of a crime scene. Please provide specific examples for each. Proper grammar, spelling, and sentence structure are expected.
Explanation:
Explanation:Those interested in working with trace evidence, such as glass, hairs, and gunshot residue, should focus on instrumentation skills and take courses in geology, soil chemistry, and materials science. If forensic biology, such as DNA analysis, is preferred, take microbiology, genetics, and biochemistry courses. Those interested in the toxicological aspects of this work, such as obtaining and interpreting toxicology reports, should study physiology, biochemistry, and chemistry.
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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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)
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
A shopper walks eastward 3.2 meters and then westward
7.2 meters.
Pro
For this motion, what is the distance moved?
"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 .
i NED HELP i’m so confused
In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a proton, assume the radius of the orbit is 5.29 10-11 m.
A) Find the magnitude of the electric force exerted on each particle.B) If this force causes the centripetal acceleration of the electron, what is the speed of the electron?
Answer:
Explanation:
Force between two charged particle having charge q₁ and q₂ at distance r
F = k q₁ x q₂ / r²
Putting the values
F = 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹ / 5.29² x 10⁻²²
= .823 x 10⁻⁷
= 823 x 10⁻¹⁰ N
This force will apply on both the charges and will be attractive in nature .
B )
Let speed of electron be v
centripetal force on electron = F
m v² / r = 823 x 10⁻¹⁰
putting the values for electron
9.1 x 10⁻³¹ x v² / 5.29 x 10⁻¹¹ = 823 x 10⁻¹⁰
v² = 478.42 x 10¹⁰
v = 21.87 x 10⁵ m /s
You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building. Before the state legislature votes on funding for the project, they would like you to prepare a report on the benefits of upgrading the elevators. One of the numbers that they have requested is the time it will take the elevator to go from the ground floor to the 102nd floor observatory. They are unlikely to approve the project unless the new elevators make the trip much faster than the old elevators. If state law mandates that elevators cannot accelerate more than 2.40 m/s2 or travel faster than 18.8 m/s , what is the minimum time in which an elevator can travel the 373 m from the ground floor to the observatory floor
Answer:
t_total = 23.757 s
Explanation:
This is a kinematics exercise.
Let's start by calculating the distance and has to reach the limit speed of
v = 18.8 m / s
v = v₀ + a t₁
the elevator starts with zero speed
v = a t₁
t₁ = v / a
t₁ = 18.8 / 2.40
t₁ = 7.833 s
in this time he runs
y₁ = v₀ t₁ + ½ a t₁²
y₁ = ½ a t₁²
y₁ = ½ 2.40 7.833²
y₁ = 73.627 m
This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.
x_total = x₁ + x₂
x₂ = x_total - x₁
x₂ = 373 - 73,627
x₂ = 299.373 m
this distance travels at constant speed,
v = x₂ / t₂
t₂ = x₂ / v
t₂ = 299.373 / 18.8
t₂ = 15.92 s
therefore the total travel time is
t_total = t₁ + t₂
t_total = 7.833 + 15.92
t_total = 23.757 s
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 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.
Kassie is 7. She completes exactly two of the 15 arithmetic problems she has been assigned for homework
before starting a video game. Kassie's behavior BEST exemplifies:
Answer: Procrastination, Defiance, Laziness
Explanation:
Answer: impulsiveness
Explanation:
A cross-country skier can complete a 7.5 km race in 45 min. What is the skier’s average speed?
Answer:
Speed = 10 km/hr
Explanation:
Speed = Distance / Time
Since we have to convert it into minutes to hours ( because it is km/hr )
45 minutes = 0.75 hours
Therefore,
Speed = 7.5 / 0.75
= 10 km/hr
Hope it helps!
Answer:
10
Explanation:
Calculate the angle θ between the radius-vector of the point and the positive x axis (measured counterclockwise from the positive x axis, within the limits of −180◦ to +180◦ ). Answer in units of ◦
The point obviously is in the 3rs quadrant
So
စ= tan^-1( y/x)-180
စ= -89.7°
The angle θ between the radius-vector of the point and the positive x axis is -89.7°
What is radius vector?Radius vector is defined as the length of the line that connects a stationary point to a moving point.
It can be defined as a line connecting a circle's center to a particle's location on its circumference.
It can also be defined as the straight line connecting two bodies moving in relative orbits, such as the path a planet takes to reach the sun at any given point along its orbit.
It is also known as position vector or location vector.
The unit vector pointing radially outward from the origin is the gradient of the position vector's length. It is typical for the spheres that are centered on the origin to have level surfaces.
Evidently, the location is in the third quadrant.
So, θ= tan⁻¹ (y / x) - 180
θ= - 89.7°
Thus, the angle θ between the radius-vector of the point and the positive x axis is -89.7°
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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.
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
The unknown vector v satisfies b - v = λ and b x v-c, where λ, b, and c are fixed and known. Find v in terms of λ, b, and c.
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