Q.468. Static Equilibrium, Application of Lami's Theorem

Question 468.
An object of mass 3kg is suspended by two light, inextensible strings. The strings make angles of 30degrees and 40degrees to the horizontal.
Find the magnitude of the tension in each spring.

Answer 468.

Refer to the figure as shown.

T = tension in the string making an angle of 30° to the horizontal

T ' = tension in the string making an angle of 40° to the horizontal

By Lami's theorem,
T/sin(90° + 40°) = T '/sin(90° + 30°) = 3 * 9.81/sin(180° - 40° - 30°)

=> T = (3 * 9.81) * cos40°/sin70° N = 24 N
and T ' = (3 * 9.81) * cos30°/sin70° N = 27.1 N.

For Lami's theorem, please refer to the following link:
Lami's Theorem

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Q.467. Friction - Minimum force to pull a sled against friction.

Question 467.

Imagine you are dragging your sibling on a sled across a flat, snowy surface at a constant speed by hauling on a rope attached to the front of the sled. The rope makes an angle θ with respect to the ground. If the coefficient of kinetic friction μk between the sled and the snow is 0.1, then what should θ be in radians if you want to exert the least amount of force necessary to keep the sled going?

 

Answer 467.

Let F = force applied
and R = normal reaction from the ground
=> Fsinθ + R = mg
=> R = mg - Fsinθ

Horizontal component of the force balances the kinetic frictional force for the sled to move with constant velocity
=> Fcosθ = (mg - Fsinθ) * 0.1
=> F (cosθ + 0.1 sinθ) = 0.1 mg

F will be minimum when cosθ + 0.1 sinθ is maximum
cosθ + 0.1 sinθ
= r [(1/r) cosθ + (0.1/r) sinθ], where r = √[1 + (0.1)^2]
= r [cos(α - θ)], where cosα = 1/r
Maximum value of cosθ + 0.1 sinθ will be for
θ = α
= arccos(1/r)
= arccos 1/√[1 + (0.1)^2]
= 5.71°.

 

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Q.466. Electrostatic equilibrium

Question 466.
Three 3.37-g Styrofoam balls of radius 2 cm are coated with carbon black to make them conducting and then are tied to 1.27-m-long threads and suspended freely from a common point. Each ball is given the same charge, q. At equilibrium, the balls form an equilateral triangle with sides of length 24.38 cm in the horizontal plane. Determine the absolute value of q.

Answer 466. Consider any one of the three balls.
Three forces act on it.
i ) its weight, mg, downwards
ii ) tension in the thread, T, in the direction from the ball
     to the point of suspension and
iii ) the resultant of the two forces of repulsion from the remaining two balls.

Let O be the point of suspension, A the position of the ball
and C the center of the equilateral triangle formed by the three balls.

OA = 1.27 m
AC cos30° = 24.38/2 cm = 0.1219 m
=> AC = (0.1219)/cos30° = 0.141 m
=> OC = √[(1.27)^2 - (0.141)^2] m = 1.262 m
=> m∠AOC = arctan[(0.141)/(1.262)] = 6.38°

Vertical component of T balances weight mg
=> Tcos6.38° = 3.37 x 10^-3 x 9.81
=> T = 0.03327 N

Horizontal component of T balances the resultant of the two forces of repulsion
from the remaining two balls
=> 0.03327sin6.38° = 2 * kq^2/(0.2438)^2 * cos30°
=> q^2 = (0.2438)^2 * 0.03327 sin6.38° / (2 * 9 x 10^9 x cos30°)
=> q^2 = 1.409669 x 10^(-14) C
=> q = 0.1187 μC.

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Q.463. de Broglie wavelength of electron in hydrogen atom

Question 463.
What is the wavelength of electron in n'th Bohr orbit ?

Answer 463.
For the electron of hydrogen atom,
mv^2/r = ke^2/r^2
=> mv^2 * r = ke^2

ByBohr's first hypothesis,
mvr = nh/2π

Taking ratio,
v = 2πke^2/nh
=> mv = 2mπ * ke^2/nh

de Broglie wavelength of the elctron
= h/mv
= nh^2 / 2 π mke^2
= 2 π n * [(1/mk) * (h/2π*e)^2]
= 2 π n * [1/(9.1 x 10^-31 x 9 x 10^9) * (6.62 x 10^-34/2π x 1.6 x 10^-19)^2 m
= 2 π n * 0.00529 x 10^8 m
= 2 π n * 0.529 angstrom units ... [1 angstrom = 10^-10 m].

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Q.461. Fluid dynamics - Bernoulli's principle

Question 461.
A liquid of density 1190 kg/m3 flows with speed 1.44m/s into a pipe of diameter 0.23m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 8.04 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.3 atm. The acceleration of gravity is 9.8 m/s2 and
Patm = 1.013 × 105 Pa.
Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe?

Answer 461.
Vb
= 1.44 * (0.23/0.05)^2
= 30.4704 m/s

According to Bernoulli's principle,
Pa + (1/2) ρVa^2 + ρgHa = Pb + (1/2) ρVb^2 + ρgHb
=> Pa - Pb
= ρ * [(1/2) (Vb^2 - Va^2) + g(Hb - Ha)]
= (1190) * [(1/2) ((30.4704)^2 - (1.44)^2) + 9.81 * 8.04]
= (1190) * (463.1858 + 78.8724) N/m^2
= 645049 N/m^2
= 645049 * 9.86923267 × 10-6 atm
= 6.366 atm
=> Pa = 6.366 + 1.3 = 7.666 atm.

For conversion of N/m^2 to atm, refer to the link as under.
Conversion Link:

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Q.460. Wave on a stretched string.

Question 460.
The mass of a string is 6.8 × 10-3 kg, and it is stretched so that the tension in it is 220 N. A transverse wave traveling on this string has a frequency of 190 Hz and a wavelength of 0.61 m. What is the length of the string?

Answer 460. Speed of the transverse wave on the string,
v = λ f
=> v = 0.61 * 190 m/s
If l = length of the string,
v = √[T/(m/l)]
=> 0.61 * 190 = √[220/(6.8 x 10^-3/l)]
=> 220 l / (6.8 x 10^-3) = (0.61 * 190)^2
=> length of the string,
l = 6.8 x 10^-3 * (0.61 * 190)^2 / 220 m
=> l = 0.4152 m = 41.52 cm.

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Q.459. Concave Mirror Problem.

Question 459.
Suppose your height is h, and you are standing in front of a concave mirror of focal length f where your real image is height h/8.73. What is your distance from the mirror in terms of f.

Answer 459.
(h/8.73)/h = v/u
=> v/u = 1/(8.73) ... ( 1 )

1/v + 1/u = 1/f
=> v = fu/(u - f)
=> v/u = f/(u - f) ... ( 2 )

From ( 1 ) and ( 2 ),
f/(u - f) = 1/(8.73)
=> u - f = 8.73f
=> u = 9.73f.

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Q.458. Circular Motion under gravity

Question 458.
A particle moves from rest at point A on the surface of a smooth circular cylinder of radius R. At B the particle leaves the cylinder. Find the equation relating θ1 and θ2.

Answer 458.
Potential energy at A = mgRcosθ1
Potential energy at B = mgRsinθ2
Gain of kinetic energy = loss of potential energy
=> (1/2) mv^2 = mgR (cosθ1 - sinθ2)
=> mv^2/R = 2mg (cosθ1 - sinθ2) ... ( 1 )

At point B, the particle will lose contact if the centripetal force = radial component of weight
=> mv^2/R = mgsinθ2 ... ( 2 )

From ( 1 ) and ( 2 ),
mgsinθ2 = 2mg (cosθ1 - sinθ2)
=> 2cosθ1 - 3sinθ2 = 0.

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Q.456. Elastic head-on collision.

Question 456.
A .45kg ice puck, moving east with a speed of 3m/s has a head on elastic collision with a .9 kg puck initially at rest. What will be the speed and direction of each object after the collision?

Answer 456.
For elastic collision, both the linear momentum as well as kinetic energies before and after collision are conserved.

Let u = velocity of the 0.45 kg ice puck
              after the collision towards the east
and v = velocity of the 0.9 kg ice puck
               after the collision towards the east

By the law of conservation of linear momentum,
initial linear momenta of the pucks before the collision
= momenta after the collision
=> (0.45) * 3 + 0.9 * 0 = 0.45 u + 0.9 v
=> u + 2v = 3 ... ( 1 )

By the law of conservation of kinetic energy,
(1/2) * (0.45) * 3^2 + 0 = (1/2) * (0.45) u^2 + (1/2) * (0.9) v^2
=> u^2 + 2v^2 = 9 ... ( 2 )

Plugging u = 3 - 2v from ( 1 ) in ( 2 ),
=> (3 - 2v)^2 + 2v^2 = 9
=> 6v^2 - 12v = 0
=> v (v - 2) = 0
=> v = 0 or 2 m/s
v cannot be zero => v = 2 m/s
and u = 3 - 2v = 3 - 2v = - 1 m/s

Answer:
Velocity of 0.45 kg puck is 1 m/s towards the west and
velocity of the 0.9 kg puck is 2 m/s towards the east.

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Q.453. Rotational Motion.

Question 453.
The moment of inertia of a hollow sphere of mass m and radius r about an axis through its centre is (2mr^2)/3. A hollow sphere is bowled across a horizontal surface. The sphere initially slides with a velocity of u. After a short interval, it starts to roll without slipping.
Show that the velocity of the ball when it starts to roll is √(3/5) u.

Answer 453.
Kinetic energy of the ball when it slides without rolling
= (1/2) mu^2
When it starts rolling without slipping, it has both linear as well as rotational motion and the above kinetic energy gets converted into kinetic energy of linear and rotational motion
If v = velocity when it rolls without slipping, then
(1/2) mu^2 = (1/2)mv^2 + (1/2) I ω^2
=> mu^2 = mv^2 + (2/3) mr^2ω^2
=> u^2 = v^2 + (2/3) v^2 ... [because ωr = v]
=> u^2 = 5/3v^2
=> v^2 = 3/5u^2
=> v = √(3/5) u.

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Q.445. Work/energy.

Question 445.
An 18 kg girl slides down a playground slide that is 3.6 m high.
When she reaches the bottom of the slide, her speed is 1.3 m/s.
a)How much energy was dissipated by friction?
b)If the slide is inclined at 20°, what is the coefficient of friction between the girl and the slide?

Answer 445.
a)

Energy dissipated by friction
= her initial PE - her final KE
= mgh - (1/2)mv^2
= 18 * [9.81 * 3.6 - 0.50 * (1.3)^2]
= 18 * (35.316 - 0.845) J
= 629.48 J

b)
Normal force = 18 * 9.81 * cos20°
=> frictional force = μ * 18 * 9.81 * cos20°
=> work done by the frictional force
= (3.6 / sin20°) * μ * 18 * 9.81 * cos20°
=> (3.6 / sin20°) * μ * 18 * 9.81 * cos20° = 629.48
=> μ = 629.48 * tan20° / (3.6 * 18 * 9.81)
=> μ = 0.36.

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Q.444. Thermal Conduction.

Question 444.
Figure (a) shows 2 metal bars X and Y which have the same cross-sectional and length joined in series. The thermal conduction of Y is three times of X.
a. If the rods are well lagged, find the temperature at the common junction.
b. The rods are then joined in parallel as shown in figure(b). Compare the rate of heat flow in figure(b) with that in figure(a) if the rods are well-lagged in both cases.

Answer 444.
a) Let T = temperature at the common junction.
Heat flow rate through both the blocks will be the same
=> Q/t = kA(T - 0)/L = 3kA(100 - T)/L
=> T = 300 - 3T
=> T = 75° C.

b)
Part a) is a case of series connection and that of b) is of parallel connection.
For part a),
Equivalent thermal resistance = L/kA + L/3kA = 4L/3kA
For part b),
equivalent thermal resistance = 1 / [1/(L/kA) + 1/(L/3kA)] = 1 / (4kA/L) = L/(4kA)
=> For a),
(Q/t)_a = 100 / (4L/3kA) = 75kA/L
and for b)
(Q/t)_b = 100 / (L/4kA) = 400kA/L
=> ratio of heat flow rate of b) to a)
[400kA/L) / (75kA/L)
= 16/3.

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Q.443. Work/Energy Problem

Question 443.
A spring with spring constant k = 195 N/m is fixed at the top of a frictionless plane inclined at an angle θ = 40°. A 1.0 kg block is projected up the plane, from an initial position that is distance d = 0.60 m from the end of the relaxed spring, with an initial kinetic energy of 15 J.
(a) What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m?
(b) With what kinetic energy must the block be projected up the plane (from its initial position) if it is to stop momentarily when it has compressed the spring by 0.40 m?

Answer 443.
(a)

Block loses kinetic energy in reaching a height (d + 0.20) sin40° which is equal to its gain of gravitational potential energy
= mg(d + 0.20) sin40° J
= 1 * 9.81 * (0.60 + 0.20) sin40° J
= 5.0446 J
Block also loses kinetic energy in doing work for compressing the spring
= (1/2) kx^2
= (1/2) * 195 * (0.20)^2 J
= 3.9 J
=> its final kinetic energy
= initial kinetic energy - kinetic energy lost in going up and compressing the spring
= 15 - 5.0446 - 3.9 J
= 6.0554 J

(b)
Kinetic energy needed to compress the spring by 0.40 m
= potential energy gained in going up (d + 0.40) sin40° + work done in compressing the spring
= mg(d + 0.40) sin40° + (1/2) k*(0.40)^2 J
= 1 * 9.81 * (0.60 + 0.40) sin40° + (1/2) * 195 * (0.40)^2 J
= 6.31 + 15.6 J
= 21.91 J.

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Q.442. Solution of electric network.

Question 442.
What is the current passing through 5 ohm resistor in given circuit ?

Answer 442.
Refer to the figure as under.

Currents, x, y and z and their combinations are shown in different branches.
For the loop ABD,
6x - 2y + 5z = 0 ... ( 1 )
For the loop ABC,
5z + 6(y+z) - 2(x-z) = 0
=> 2x - 6y - 13z = 0 ... ( 2 )

For the loop BCD,
6(y+z) + 2y + 2(x+y) = 10
=> 2x + 10y + 6z = 10 ... ( 3 )

Solving eqns. ( 1 ), ( 2 ) and ( 3 ) using 
Wolfram Alpha Link
z = - 0.4 A
=> current through 5 ohm resistor is 0.4 A.

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Q.440. Static equilibrium.

Question 440.
The framework is supported by the member AB which rests on a smooth floor.
When loaded the pressure distribution on AB is linear as shown in the figure. Determine the length d of member AB and the intensity w for this case.

Answer 440.
Let a = bottom surface area of the beam AB.
First we find the moment of the force distribution wa about A and then equate it to the moment of 800 lb. Consider a small element dx of AB at a distance x
Force on it,
dF = wa * (x/d) dx
Moment of this force about A
dM = (wa/d) x^2 dx
Total moment found by integrating between limits x = 0 to x = d is
(wa/d) ∫ x^2 dx ... (x = 0 to x = d)
= wad^2/3
Equatting to the moment of force 800 lb
=> wad^2/3 = 800 * 4
=> wad^2 = 9600 ... ( 1 )

Equatting the force in the vertical direction,
800 = (wa/2) d
=> wad = 1600 ... ( 2 )

Solving ( 1 ) and ( 2 ),
d = 6 ft and
wa = 1600/6 lb = 267 lb.

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Q.438. Static equilibrium, Couples of forces.

Question 437.
If the resultant couple of the three couples acting on the triangular block is to be zero, determine
the magnitude of forces F and P.
Given:    F₁ = 150 N,   a= 300 mm,   b = 400 mm,   d= 600 mm.

Answer 437.
Couple Pd acts downwards and couple F1b acts upwards
Balancing, Pd = F1b
=> P = F1 * (b/d) = 150 * (400/600) = 100 N.

Couple Fd acts to the left and F1a acts to the right
Balancing, Fd = F1a
=> F = F1 (a/d) = 150 * (300/600) = 75 N.

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Q. 437. Static Equilibrium involving forces and moments

Question 437.
The forces and couple moments which are exerted on the toe and heel plates of a snow ski are given by
F_t, M_t and F_h, M_h respectively. Replace this system by an equivalent force and couple moment acting at point P. Express the results in Cartesian vector form.
Given:  a= 120 mm,   b= 800 mm,  
           Ft = -50i + 80j - 158k N,   Mt = -6i + 4j + 2k Nm  and
           Fh = -20i + 60j -250k N,   Mh = -20i + 8j + 3k Nm.

Answer 437.
Moment due to F_h
= (b, 0, 0) x (-20, 60, -250)
= (0.8, 0, 0) x (-20, 60, -250)
= (0, 200, 48) Nm

Moment due to F_t
= (a + b, 0, 0) x (-50, 80, -158)
= (0.92, 0, 0) x (-50, 80, - 158)
= (0, 145.36, 73.6) Nm

Sum of moments due to F_h and F_t
M = (0, 345.36, 121.6)
Adding Mt and Mh
total moment
= (0 - 6 - 20, 4 + 8 + 345.36, 2 + 3 + 121.6)
= (-26, 357.36, 126.6)
≈ 0- 26i + 357j + 127k Nm.

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Q.436. Centre of area

Question 436.
Locate the centroid (x_c, y_c)
of the shaded area?
Given:  a= 1 in,   b= 6 in,   c= 3 in,   d= 3 in

Answer 436. Refer to the link: http://en.wikipedia.org/wiki/List_of_cen…

Area of triangle = (1/2) cd
Center of triangle = (b + c/3, d/3)

Area of rectangle = bd
Center of rectangle = (b/2, d/2)

Area of quarter circle = πd^2/4
Center of quarter circle = ( - 4d/3π, 4d/3π)

Area of semicircular hole = πa^2/2
Center of semicircular hole = (0, 4a/3π)

=> x_c
= [(1/2)cd * (d/3) + bd * d/2 + (πd^2/4) * (- 4d/3π) - (πa^2/2) * 0] / [(1/2)cd + bd + πd^2/4 - πa^2/2]
= [(9/2) * 7 + 18 * 3 + (7.0685775) * (- 1.27324) - (1.5708) *0] / [(9/2) + 18 + (7.0685775) - (1.5708)]
= (31.5 + 54 - 9 ) / (4.5 + 18 - 5.5)
= (76.5) / (28)
= 2.732.
and y_c
= [(1/2)cd * (b + c/3) + bd * b/2 - (πd^2/4) * (4d/3π) - (πa^2/2) * (4a/3π)] / [(1/2)cd + bd + πd^2/4 - πa^2/2]
= [(9/2) * 1 + 18 * 1.5 + (7.0685775) * (1.27324) - (1.5708) * (0.42441)] / [(9/2) + 18 + (7.0685775) - (1.5708)]
= (4.5 + 27 + 9 - 0.67) / (4.5 + 18 - 5.5)
= (39.83) / (28)
= 1.423.

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Q.432. Vertical motion under gravity.

Question 432.
An object is thrown downward with an initial speed of 6 m/s from a height of 42 m above the ground. At the same instant, a second object is propelled vertically up from ground level with a speed of 55 m/s. At what height above the ground will the two objects pass each other?

Answer 432.
Let h = height above the ground where they pass each other.
=> 42 - h = 6t + 4.95t^2
and h = 55t - 4.95t^2
Adding,
42 = 61 t
=> t = 42/61
Plugging this value of t in h = 55t - 4.95t^2
=> h
= 55 * (42/61) - 4.95 * (42/61)^2
= 37.87 - 2.35
= 35.52 m.

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Q. 428 Vertical motion under gravity.

Question 428.
A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.70 m/s^2 . At 10.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.
How long after it was launched will the rocket fall back to the launch pad?

Answer 428.
The upward velocity of the rocket after 10 s = 2.70 * 10 = 27 m/s
Distance traveled in 10 s = (1/2) * 2.7 * 10^2 = 135 m

Let t = time for the rocket to reach the ground from this height of 135 m
Using s = ut + (1/2) gt^2
with s = - 135, u = 27, g = - 9.81
=> - 135 = 27t - 4.95 t^2
=> 4.95 t^2 - 27t - 135 = 0
=> t = (1/9.81) [27 + √((27)^2 + 2673)]
=> t = 8.7 s
=> time to fall back after launching
= 10 + 8.7 s
= 18.7 s.

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