Sabtu, 30 September 2017
APPLE has rolled-out its first update to iOS 11, which launched worldwide last week. The new software update includes a number of fixes for issues encountered by those who have already upgraded.
Kamis, 28 September 2017
Up for Auction: Einstein Letters Detailing General Relativity and Grand Unified Theory..
Two letters written by Albert Einstein detailing his thoughts on some of the most famous theories in physics are going up for auction this week, according to Nate D. Sanders Auctions, which is auctioning off the letters.
Einstein wrote the two letters to his friend and scientific sounding board Michele Besso. One letter outlines Einstein's early thoughts on a grand unified theory, while the other, written much later, is a bit more philosophical, asking what can be known and whether some physical theories can ever be proved.
The bidding for each letter starts at $60,000. Sam Heller, a spokesman for the auction house, said the letters came from a private collection, whose owner didn't want to be identified. [8 Ways You Can See Einstein's Theory of Relativity in Real Life]
In one of the letters, written by Einstein in Gatow, a district of Berlin, Germany, on Sept. 5, 1929, he detailed his thoughts on the so-called grand unified theory in physics. At the time, Einstein was in Gatow attempting to escape the rush of publicity — the press and thousands of well-wishers — surrounding his 50th birthday; he was staying with a friend and physician named Janos Plesch, according to Albrecht Fölsing's biography of Einstein.
Albert Einstein wrote this letter about the grand unified theory on Sept. 5, 1929, while he was in Gatow, a district of Berlin, Germany.Credit: Nate D. Sanders Auctions
At that point in his life, Einstein was working on unifying gravity and electromagnetism, as he felt that both were underlying manifestations of a deeper theory. He didn't succeed — physicists still haven't managed to do it — but the letter illustrates the directions he thought might be fruitful for solving the theory. Notably, he dispensed with parts of quantum mechanics, which describes the behaviour of particles in a probabilistic way, as illustrated in one part of the letter:
"However, the very best thing, on which I have worked for days and half the nights, speculating and making calculations, is now completed and lying in front of me, compressed into seven pages with the title 'Unified Field Theory.' It looks antiquated, and the dear colleagues, including you, my dear, will initially stick their tongues out as far as possible. After all, these equations do not contain Planck's constant h." [Einstein Quiz: Test Your Knowledge of the Famous Genius]
Planck's constant, which determines the relationship between energy and wavelength, appears in most quantum mechanical theories, and in the late 1920s, Einstein wasn't as enamoured of the theory as many of his physicist colleagues were; he felt that the statistical nature of quantum mechanics was a problem for the theory. In other words, in many quantum mechanical applications, you can find a probability that a particle will be in a given place, for example, but you can't say for sure exactly where it is. Einstein thought that meant the theory was incomplete.
Einstein says as much in the letter: "But once they have clearly reached the performance limit of the statistics craze, people will remorsefully return to the time-space concept, and then these equations will constitute a point of departure." Basically, he thought that his work might be a first step in finding a unified theory not based on probabilistic calculations, the way quantum mechanics is, but rather something more akin to classical theories like mechanics, where, for example, the location of an object can be specified exactly.
Einstein claimed to have found a way to include a "distant parallelism" when describing space-time. Distant parallelism, which states that parallel lines never meet, is a basic premise of the Euclidean geometry most people learn in school. But Euclidean geometry doesn't always occur in geometries used in some general relativistic calculations. That particular methodology ended up not producing a unified theory.
The second letter was written much later, on April 15, 1950. By that time, Einstein was in Princeton at the Institute for Advanced Study, where he would remain until his death in 1955. This letter also talks about general relativity. One sentence seems to speak of a relationship to God: "There is one thing that I have learned in the course of a long life: It is devilishly difficult to get closer to 'Him' if one does not want to remain on the surface."
But to say Einstein was speaking of God would be a misreading, said Galina Weinstein, a historian of science who authored "Einstein's Path to the Special Theory of Relativity" (Cambridge Scholars Publishing, 2017). Weinstein said the "Him" reference has little to do with religion.
"Einstein does not talk about what connections with God he had (or didn't have). He only talks about unified field theory," she told Live Science in an email. The rest of the paragraph bears this out; the discussion is about his mathematical technique for coming up with something like a unified theory that could reconcile gravity with both electromagnetism and quantum mechanics. (That, too, proved to be a dead end.)
The Nate Sanders auction house has sold other Einstein memorabilia, Heller said. In August, the firm sold a 1938 letter from Einstein to Bessolamenting the agreement British Prime Minister Neville Chamberlain signed with Germany for $31,250, and an autographed copy of a famous photo of Einstein sticking out his tongue sold for $125,000 in July.
Rabu, 27 September 2017
MCQs on Friction
Friction
Q1. The force of friction acts in a direction _____ to the direction of motion of the object.
a. Same.
b. Opposite👈
c. Perpendicular.
d. Downwards
Q2. The force of friction depends upon
a. Nature of surface of contact.
b. The material of objects in contact.
c. Both ‘a’ and ‘b’👈
d. None of the above
Q3. The body will move only when
a. Force of friction = applied force.
b. The force of friction < applied force👈
c. Force of friction > applied force.
d. All of the above
Q4. The ratio of the limiting force of friction (F) to the normal reaction (R) is known as
a. Coefficient of friction👈
b. Force of friction.
c. The angle of friction.
d. None of the above
Q5. The coefficient of friction (µ) is equal to?
a. TanΦ👈
b. SinΦ.
c. CotΦ.
d. CosΦ
(Where Φ = angle of friction)
Q6. The force of friction (F) is equal to
a. µR/2
b. µR👈
c. 2µR
d. µR/3
Q7. The value of Normal reaction (R) for the following figure is
a. W – PSinθ👈
b. W + PSinθ
c. P – WSinθ
d. P + WSinθ
(Where, W = Weight of block, P = Applied force, µ = Coefficient of friction, θ = Angle)
Q8. When the two surfaces in contact have a thick layer of lubricant in between them, it is known as
a. Solid friction.
b. Rolling friction.
c. Greasy friction.
d. Film friction👈
Q9. When the two surfaces in contact have a very thin layer of lubricant in between them, it is known as
a. Solid friction.
b. Rolling friction.
c. Greasy friction.
d. Film friction👈
Q10. The force of friction is maximum when the surface
Q10. The force of friction is maximum when the surface
a. Is on the point of motion👈
b. Is at rest
c. Is moving
d. The friction remains same at all points
b. Is at rest
c. Is moving
d. The friction remains same at all points
Q11. The types of threads used in Screw jacks are
a. Metric thread
b. Square👈
c. ACME
d. Buttress
b. Square👈
c. ACME
d. Buttress
Q12. For a single started thread, the lead is equal to
a. p/2
b. p👈
c. 2p
d. p/3
Where p = pitch
Q13. The slope of the thread with horizontal is known as
a. helix
b. lead
c. pitch
d. helix angle👈
Q14. The flat pivot bearing is used to bear
a. axial thrust👈
b. radial thrust
c. both radial and axial thrust
d. None of the above
Q15. In plate clutch, the clutch plate is placed
a. Before flywheel
b. After pressure plate
c. In between pressure plate and flywheel👈
d. None of the above
Q16. The clutch is placed in the four-wheeler, in between
Engine and gearbox👈
Gear box and differential
Propeller shaft and differential
None of the above
Q17. The radius of the friction circle is independent of
a. Weight of the shaft👈
b. Radius of the shaft
c. Coefficient of friction
d. All of the above
Q18. In rotation of shaft, the power lost in friction is given by
a. W x µ x V Watts👈
b. W x µ x V Kilo-watts
c. W x µ / V Watts
d. W x µ / V Kilo-watts
Where, W=Weight of shaft, µ=Coefficient of friction, V= Velocity of shaft (in m/s)
Senin, 25 September 2017
MCQs on Friction
Friction
Q1. The force of friction acts in a direction _____ to the direction of motion of the object.
a. Same.
b. Opposite👈
c. Perpendicular.
d. Downwards
Q2. The force of friction depends upon
a. Nature of surface of contact.
b. The material of objects in contact.
c. Both ‘a’ and ‘b’👈
d. None of the above
Q3. The body will move only when
a. Force of friction = applied force.
b. The force of friction < applied force👈
c. Force of friction > applied force.
d. All of the above
Q4. The ratio of the limiting force of friction (F) to the normal reaction (R) is known as
a. Coefficient of friction👈
b. Force of friction.
c. The angle of friction.
d. None of the above
Q5. The coefficient of friction (µ) is equal to?
a. TanΦ👈
b. SinΦ.
c. CotΦ.
d. CosΦ
(Where Φ = angle of friction)
Q6. The force of friction (F) is equal to
a. µR/2
b. µR👈
c. 2µR
d. µR/3
Q7. The value of Normal reaction (R) for the following figure is
a. W – PSinθ👈
b. W + PSinθ
c. P – WSinθ
d. P + WSinθ
(Where, W = Weight of block, P = Applied force, µ = Coefficient of friction, θ = Angle)
Q8. When the two surfaces in contact have a thick layer of lubricant in between them, it is known as
a. Solid friction.
b. Rolling friction.
c. Greasy friction.
d. Film friction👈
Q9. When the two surfaces in contact have a very thin layer of lubricant in between them, it is known as
a. Solid friction.
b. Rolling friction.
c. Greasy friction.
d. Film friction👈
Q10. The force of friction is maximum when the surface
Q10. The force of friction is maximum when the surface
a. Is on the point of motion👈
b. Is at rest
c. Is moving
d. The friction remains same at all points
b. Is at rest
c. Is moving
d. The friction remains same at all points
Q11. The types of threads used in Screw jacks are
a. Metric thread
b. Square👈
c. ACME
d. Buttress
b. Square👈
c. ACME
d. Buttress
Q12. For a single started thread, the lead is equal to
a. p/2
b. p👈
c. 2p
d. p/3
Where p = pitch
Q13. The slope of the thread with horizontal is known as
a. helix
b. lead
c. pitch
d. helix angle👈
Q14. The flat pivot bearing is used to bear
a. axial thrust👈
b. radial thrust
c. both radial and axial thrust
d. None of the above
Q15. In plate clutch, the clutch plate is placed
a. Before flywheel
b. After pressure plate
c. In between pressure plate and flywheel👈
d. None of the above
Q16. The clutch is placed in the four-wheeler, in between
Engine and gearbox👈
Gear box and differential
Propeller shaft and differential
None of the above
Q17. The radius of the friction circle is independent of
a. Weight of the shaft👈
b. Radius of the shaft
c. Coefficient of friction
d. All of the above
Q18. In rotation of shaft, the power lost in friction is given by
a. W x µ x V Watts👈
b. W x µ x V Kilo-watts
c. W x µ / V Watts
d. W x µ / V Kilo-watts
Where, W=Weight of shaft, µ=Coefficient of friction, V= Velocity of shaft (in m/s)
Minggu, 24 September 2017
Will this be a difference we can see in the sheetmetal of upcoming Ford models?
Don’t lie, When asked which superpower you’d most want to have, it’d be money. Unlike broke and humble Peter Parker or muscled-up Superman, superheroes like Batman get their powers by spending billions on next-generation technology. Better still is that they can taste the good life and, when their consciouses question the morality behind being greenback gluttons, they can justify it by the fact they use the money to save lives and fight crime. Of course, the more fun of the two main monied superheroes is Iron Man.
He lives in Malibu, California and not depressing Gotham and uses holographic devices to develop his super suits. Taking a cue from him, Ford has recently decided to team up with Microsoft to use the latest in hologram technology to develop the cars of the future. The technology is called HoloLens and is essentially a pair of wireless goggles that use augmented reality to paint a picture of sorts onto a physical object. In this case, that object is a full-scale clay model of a future model like the Ford Fusion. Ford recently let it be known that it’s using the technology to augment the old school clay model process by shortening the time it takes to see physical changes on the model.
“It’s amazing we can combine the old and the new – clay models and holograms – in a way that both saves time and allows designers to experiment and iterate quickly to dream up even more stylish, clever vehicles,” says Jim Holland, Ford vice president, vehicle component and systems engineering. “Microsoft HoloLens is a powerful tool for designers as we continue to re-imagine vehicles and mobility experiences in fast-changing times.” This could prove to be a boon for designers, who now have more tools at their fingertips and more time freed up to welcome in the spirit of inspiration. For example, it used to take days to manifest proposed changes to a grille on a clay model.
Using Microsoft’s HoloLens technology, designers can simply slap on a pair of goggles and see it as if it were already on the model. Now if only they’d use it to fix the Taurus and C-Max
Kamis, 14 September 2017
Resolution of forces
Mechanics
It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics to design, taking into account the effects of forces. Statics deal with the condition of equilibrium of bodies acted upon by forces.
It is defined as that branch of science, which describes and predicts the conditions of rest or motion of bodies under the action of forces. Engineering mechanics applies the principle of mechanics to design, taking into account the effects of forces. Statics deal with the condition of equilibrium of bodies acted upon by forces.
Rigid bodyA rigid body is defined as a definite quantity of matter, the parts of which are fixed in position relative to each other. Physical bodies are never absolutely but deform slightly under the action of loads. If the deformation is negligible as compared to its size, the body is termed as rigid.
Force
Force may be defined as any action that tends to change the state of rest or motion of a body to which it is applied.
The three quantities required to completely define force are called its specification or characteristics. So the characteristics of a force are:
1. Magnitude
2. Point of application
3. Direction of application
The line of action of force
The direction of a force in the direction, along with a straight line through its point of application in which the force tends to move a body when it is applied. This line is called the line of action of force.Representation of force
Graphically a force may be represented by the segment of a straight line.
Various Systems of Forces
Parallel forces on a plane
1. Like parallel forces: Coplanar parallel forces when acting in the same direction.
2. Unlike parallel forces: Coplanar parallel forces when acting in opposite direction.
3. Resultant of like parallel forces: Let P and Q are two like parallel forces act at points A and B.
R = P + Q 
4. Resultant of unlike parallel forces:
R = P - Q 
Couple Force
Two, unlike equal parallel forces, form a couple.
The rotational effect of a couple is measured by its moment. Moment = P × l
Sign convention: Anticlockwise couple (Positive)
Clockwise couple (Negative)
The rotational effect of a couple is measured by its moment. Moment = P × l
Sign convention: Anticlockwise couple (Positive)
Clockwise couple (Negative)
Question 1: A rigid bar CABD supported as shown in the figure is acted upon by two equal horizontal forces P applied at C and D. Calculate the reactions that will be induced at the points of support. Assume l = 1.2 m, a = 0.9 m, b =0.6 m.
Solution: Taking moment about A,
Ra = Rb
Rb ✖l + P ✖ b = P ✖ a
Rb = P(0.9 - 0.6) ➗ 1.2
Rb = 0.25P(↑)
Ra = 0.25P(↓)
Question2: Owing to weight W of the locomotive shown in the figure, the reactions at the two points of support A and B will each be equal to W/2. When the locomotive is pulling the train and the drawbar pull P is just equal to the total friction at the points of contact A and B, determine the magnitudes of the vertical reactions Ra and Rb.
Solution
åV = 0Ra + Rb = W
Taking moment about B,
å MB = 0
Ra ´ 2a + P ´ b = W ´ a
Þ Ra =(W .a - P.b)
2a
Solution
åV = 0
Ra + Rb = 3P + Q
Þ Ra + Rb = 3´ 90 + 72
Þ Ra + Rb = 342KN
å M A = 0
Rb ´ 9.6 = 90 ´1.8 + 90 ´ 3.6 + 90 ´ 5.4 + 72 ´ 8.4
Þ Rb = 164.25KN
\ Ra = 177.75KN
Question 4: The beam AB in figure is hinged at A and supported at B by a vertical cord which passes over a friction-less pulley at C and carries at its end a load P. Determine the distance x from A at which a load Q must be placed on the beam if it is to remain in equilibrium in a horizontal position. Neglect the weight of the beam.
Solution
Take å M A = 0
S ´ l = Q ´ x
Þ x = P.l Q
Composition of force
The reduction of a given system of forces to the simplest system that will be its equivalent is called the problem of the composition of forces into components.
These components are also referred to as rectangular components of a force.
To find the component of a vector along a given axis, we drop a perpendicular on the given axis from the vector.
For example, OA is the given vector. We have to find its component along the horizontal axis. Let us call it the x-axis. We drop a perpendicular AB from An onto the x-axis. The length OB is the component of OA along the x-axis. If OA makes angle p with the horizontal axis, then in triangle OAB, OB/OA = Cos P or OB = OA Cos P.
Remember that component of a vector is a scalar quantity. If the component is along the negative direction, we put a (-) sign with it.)
Usually, we resolve the vector into components along mutually perpendicular components.
OB is the x component OB = OA Cos p.
Similarly component along the vertical direction or the y-axis is OC
OCAB is a rectangle.
So OC = AB
look at triangle OAB again,
AB/OA = Sin p
=> AB = OA Sin p = OC
Thus y component OC = OA Sin p.
Note that p is the angle with the horizontal axis.
Solved Numericals on "Rectangular components of a Force"
Question
Find the X and Y components of a 25 m displacement at an angle of 210 deg.
Answer
OA is the displacement vector.
The angle with the horizontal axis is 210 deg - 180 deg = 30 deg
x component = OB = -25 Cos 30 deg = -21.7
y component = AB = -25 Sin 30 deg = -12.5 m
Note that each component is pointing along the negative coordinate direction and thus we must take it as negative.
Now we will solve a problem using the component method
Question
Find the resultant of the following two displacements: 2 m at 30 deg and 4 m at 120 deg. The angles are taken relative to the x-axis.
Answer
Rx = 2 Cos 30 deg - 4 Cos 60 deg = - 0.268 m
Ry = 2 Sin 30 deg + 4 Sin 60 degg = 4.46 m
R = √(Rx2 + Ry2)
= √(-0.2682 + 4.462) = 4.47 m
tan q = Ry/Rx = 4.46/0.268
=> q = 86.6 deg
p = 180 deg - 86.6 deg = 93.4 deg
Parallelogram law
If two forces represented by vectors OA and OB acting under an angle α are applied to a body at point O. Their action is equivalent to the action of one force, represented by vector OC, obtained as the diagonal of the parallelogram constructed on the vectors OA and OB directed as shown in the figure.
Force OC is called the resultant of OA and OB and the forces are called its components.
Force OC is called the resultant of OA and OB and the forces are called its components.
Special cases
Question
Two forces of 3 N and 4 N are acting at a point such that the angle between them is 60 degrees. Find the resultant force
Answer
Magnitude R of the resultant force is R = √(32 + 42 + 2 x 3 x 4 Cos 60 deg)
= √(9 + 16 + 12) = √(37 = 6.08 N
Direction of R is given by finding the angle q
tan q = (3 Sin 60 deg)/(4 + 3 Cos 60 deg) = 0.472
q = tan-1 0.472
= 25.3 deg
Thus R is 6.08 N in magnitude and is at an angle of 25.3 deg to the 4 N force.
Question
A car goes 5 km east 3 km south, 2 km west and 1 km north. Find the resultant displacement.
Answer
First, we will make the vector diagram
O to A 5 km east
A to B 3 km south
B to C 2 km west
c to D 1 km north
Net displacement is OD
Along the horizontal direction: 5 km east - 2 km west = 3 km east
Along the vertical direction: 3 km south - 1 km north = 2 km south
OD = √(32 + 22 + 2 x 3 x Cos 90 deg)
= √(32 + 22)
= 3.6 km
tan p = 2/3
or p = tan-12/3 = 34 deg
Thus resultant displacement is 3.6 km, 34 deg south of east.
Action and reaction
Often bodies in equilibrium are constrained to investigate the conditions.
Free body diagram
Free body diagram is necessary to investigate the condition of equilibrium of a body or system. While drawing the free body diagram all the supports of the body are removed and replaced with the reaction forces acting on it.
Draw the free body diagrams of the following figures.
Draw the free body diagrams of the following figures.
 |
3. Draw the free body diagram of the following figures. |
Equilibrium of colinear forces:
Two forces can be in equilibrium only if they are equal in magnitude, opposite in direction and collinear in action.
Figure: Tension
Figure: Compression
Equilibrium of concurrent forces in a plane
- If a body known to be in equilibrium is acted upon by several concurrent, coplanar forces, then these forces or rather their free vectors, when geometrically added must form a closed polygon.
- This system represents the condition of equilibrium for any system of concurrent forces in a plane.
Ra w tan S w sec
Lami’s theorem
If three concurrent forces are acting on a body kept in an equilibrium, then each force is proportional to the sine of the angle between the other two forces and the constant of proportionality is same.
Problem: An electric light fixture of weight Q = 178 N is supported as shown in the figure. Determine the tensile forces S1 and S2 in the wires BA and BC, if their angles of inclination are given.
Solution:
Find S1 and S2 to get Answer.
Theory of transmissibility of a force
The point of application of a force may be transmitted along its line of action without changing the effect of force on any rigid body to which it may be applied.
Moment of a Force
- Considering wrench subjected to two forces P and Q of equal magnitude. It is evident that force P will be more effective compared to Q, though they are of equal magnitude.
- The effectiveness of the force as regards it is the tendency to produce rotation of a body about a fixed point is called the moment of the force with respect to that point.
- Moment = Magnitude of the force × Perpendicular distance of the line of action of force.
- Point O is called the moment centre and the perpendicular distance (i.e. OD) is called moment arm.
- Unit is N.m
Theorem of Varignon
The moment of the resultant of two concurrent forces with respect to a centre in their plane is equal to the algebraic sum of the moments of the components with respect to some centre.
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