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.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.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.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.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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Q.427. Vertical motion under gravity.

Question 427.
A Rubber ball is dropped from a height of 27m.; Each time that it hits the ground it bounces to a height 2/3 of that from which it fell. Find the total distance travelled by the ball until it comes to rest?

Answer 427.
Total distance = 27 + (2/3) * 27 + (2/3) * 27 + (2/3)^2 * 27 + (2/3)^2 * 27 + ...
= 27 [1 + 2*(2/3) + 2 * (2/3)^2 + ...]
= 27 [1 + 2*(2/3) {1 + (2/3) + (2/3)^2 + ...}]
= 27 [1 + 2*(2/3) * 1/(1 - 2/3)]
= 27 [1 + 4/3 * (3/1)]
= 135 m.

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Q.424. Application of Bernoulli's equation.

Question 424.
Use Bernoulli's equation to estimate the upward force on an airplane's wing if the average flow of speed of air is 191 m/s ABOVE the wing and 162 m/s BELOW the wing. The density of the air is 1.6 kg/m3 and the area of each wing surface is 26.3m3.

Answer 424.
Pressure below the wing is more than pressure above the wing
=> P1 + (1/2) dv1^2 = P2 + (1/2) dv2^2
=> P2 - P1
= (1/2) * 1.6 * [(191)^2 - (162)^2]
= 0.8 * 10237 N/m^2
= 8189.6 N/m^2
=> upward force
= (P2 - P1) x area
= 8189.6 * 26.3 N
= 215386.5 N.

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Q.418. Gravitation.

Question 418.
The radius of the earth is R and its angular velocity is W (omega) . If the earth suddenly stops rotation, then the change in the value of g at a place on the equator will be :
A- zero          B-RW          C- R/W^2          D- RW^2.

Answer 418.
g is the acceleration experienced by the body on the surface of the earth.
The body of mass m on the surface of the earth exerts a force on the surface equal to its weight, mg and has a reaction from the surface of the earth = N
The net force on the body of mass m = mg - N provides the centripetal force needed for the revolution of the object on the surface on the circle of radius R with angular velocity W
=> mg - N = mRW^2
=> N = mg - mRW^2
N is the weight recorded on the balance = mg'
=> mg' = mg - mRW^2
=> m(g - g') = RW^2
=> g - g' = RW^2 which is the difference in acceleration due to gravity when the earth is rotating and when it stops rotating.

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Q.415. Mechanics (Statics - Parallel forces).

Question 415.
A 4.4 m uniform pipe whose weight is 450 N is supported at each end by cords. two loads of 327N and 396 N are suspended 1.5m from the left end and 2.8m from the right end respectively. Find the tension in each cord.

Answer 415.
Let T1 and T2 be the tensions in the left and right cord respectively.

Taking moment at the right end of the rod,
T1 * 4.4 = 327 * (4.4 - 1.5) + 450 * 2.2 + 396 * 2.8
=> 4.4 T1 = 948.3 + 990 + 1108.8
=> T1 = 692.5 N.

Similarly, taking moment about the left end of the rod,
T2 * 4.4 = 450 * 2.2 + 396 * (4.4 - 2.8) + 327 * 1.5
=> 4.4 T2 = 990 + 633.6 + 490.5
=> T2 = 480.5 N

======================================
Verification:
Total upward forces
= T1 + T2 = 692.5 + 480.5 = 1173 N.
Total downward forces
= 450 + 327 + 396 = 1173 N.

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Q.414. Projectile Motion

Question 414.
In a game of rugby, a penalty kick is taken. The formula H(x) = 0.889x - 0.0198x^2 models the flight of the rugby ball. H(x) measures the height of the ball above the ground (in metres) at a distance x metres along a straight line on the ground beginning at the point where the penalty kick is taken and extending beyond the two goal posts.
i) How far from where the ball was kicked did it land (nearest meter)?
ii) Calculate the maximum height H reached by the ball (nearest meter).
iii) After the kick is taken, it just clears the cross-bar by 1cm. How far away from the goal posts does the formula suggest the penalty kick was taken? Set up the equation to be solved, and the equation that will help you solve it. Round sensibly.
Note: The height of the cross-bar is 3 m above the ground.

Answer 414.
i)
H = 0 where the ball lands
=> 0 = 0.889x - 0.0198x^2
=> 0 = 0.889 - 0.0198x ... [because x is not zero where it lands]
=> x = 45 m.

ii)
Maximum height is reached at x = (1/2) * (0.889/0.0198)
=> H = 0.889 * (1/2) * (0.889/0.0198) - (0.0198) * (1/2)^2 * (0.889/0.0198)^2
=> H = (1/2) (0.889)^2/(0.0198) m
=> H = 20 m.

iii)
Plugging H = 3.01
=> 3.01 = 0.889x - 0.0198x^2
=> x^2 - (0.889/0.0198)x + (3.01/0.0198) = 0
=> x^2 - 44.899x + 152.02 = 0
=> x = (1/2) [44.899 ±√[2015.92 - 608.08]
=> x = (1/2) (44.899 ± 37.521)
=> x = 3.689 m or 41.21 m.
41.21 m is the likely answer.

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Q.412. Mechanics - Pulley.

Question 412.
A cord is padded over a weightless and frictionless pulley.  Masses of 50 grams and 60 grams are attached to the ends of the cord. Determine the acceleration of the system and find the distance the masses will move during the third second after they are stared from rest.

Answer 412.
Let T = tension in the cord
=> 0.06 * 9.81 - T = 0.06 a
and T - 0.05 * 9.81 = 0.05 a

Adding,
0.01 * 9.81 = 0.11 a
=> a = 0.892 m/s^2

Distance travelled in the third second
= S3 - S2
= (1/2) a (3^2 - 2^2)
= (1/2) * (0.892) * 5 m
= 2.23 m.

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Q.411. Gravitation

Question 411.
A spacecraft lands on a newly discovered planet orbiting the star Antares.To measure the acceleration due to gravity on the planet,Dave ,an astronaut, drops a rock from a height of 2m. A precision timer indicates that it takes the rock 0.71s to fall to the ground.
a-Calculate the acceleration due to gravity on the new planet.
b-Find the mass of the new planet ,given that the radius of the planet is 6.05*10^6m.
c-Calculate the gravitional accelration at the location 50.0km from the surface of this planet.

Answer 411.
a)
2 = (1/2) a * (0.71)^2
=> a = 4/(0.71)^2 = 7.935 m/s^2

b)
a = GM/R^2
=> Mass, M
= aR^2/G
= (7.935) * (6.05 * 10^6)^2 / (6.67 * 10^-11) kg
= 4.35 x 10^24 kg

c)
acceleration at the height, h of 50 km
= GM/(R + h)^2
= 6.67 x 10^-11 * 4.35 x 10^24 / [(6.05 + 0.05) x 10^-6]^2 m/s^2
= 7.80 m/s^2.

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