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.408. Motion under gravity

Question 408.
Let a b c d be points on a vertical line such that ab = bc = cd if a body is released from the position a, find the ratio of the times of descent through ab bc and cd.

Answer 408.
Let ab = bc = cd = x
=> x = (1/2)gt1^2
2x = (1/2)gt2^2
3x = (1/2)gt3^2
=>
t1 = √(2x/g)
t2 = √(4x/g) and
t3 = √(6x/g)
=>
ratio of times of descent through ab, bc and cd
= t1 : (t2 - t1) : (t3 - t2)
= √(2x/g) : √(4x/g) - √(2x/g) : √(6x/g) - √(4x/g)
= √2 : (2 - √2) : (√6 - 2).

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Q. 401. Rotational motion/tension problem

Question 401.
A 11 kg solid steel cylinder with a 10 cm radius is mounted on bearings so that it rotates freely about a horizontal axis. Around the cylinder is wound a number of turns of a fine gold thread. A 1.0 kg monkey named Fred holds on to the loose end and descends on the unwinding thread as the cylinder turns.
Compute Fred's acceleration (m/(s^2)) and the tension in the thread (N).

Answer 401.
Net force on the monkey,
mg - T = ma, where a = acceleration of the monkey and T = tension in the thread
=> 9.81 - T = a ... ( 1 )

Torque acting on the cylinder w.r.t. to its axis
= (moment of inertia of the cylinder) * (its angular acceleration) = 0.1 T
=> (1/2) * 11 * (0.10)^2 * α = 0.1 T
Also, α * 0.1 = a
=> (1/2) * 11 * (0.10)^2 * a/(0.10) = 0.1 T
=> 5.5a = T ... ( 2 )

Solving ( 1 ) and ( 2 ),
a = 9.81/6.5 = 1.51 m/s^2
T = 5.5a = 5.5 * 1.51 N = 8.376 N.

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Q.109. Projectile Motion.

Question 109.
A soccer player kicks a rock horizontally off a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.02 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

Answer 109.
Since the player kicked the ball horizontally, vertical velocity of the ball is zero.
Time taken to go down by 40 m with zero initial velocity = t
=> 40 = (1/2)gt^2
=> t = √(80/9.81) = 2.86 s.

=> time taken by sound to reach the player
= 3.02 - 2.86 s
= 0.16 s
and distance covered by sound
= 343 * 0.16 m
= 54.88 m

=> horizontal distance covered by the ball
= √[(54.88)^2 - (40)^2] m
= 37.6 m.
This distance was covered by the ball moving at constant horizontal velocity, v, in the same time as it took to reach down to the surface of water, i.e., 2.86 s.
=> v
= 37.6/2.86 m/s
= 13.15 m/s.

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Q.46. Length of a wire wrapped around a cylinder.

Question 46.

Suppose that eight turns of a wire are wrapped around a pipe with length of 20 cm and a circumference of 6 cm. What is the length of the wire?

Answer 46.

If you cut the cylinder along a line parallel to its axis and open up it will be a rectangle of length 20 cm and width 6 cm. As there are 8 turns, pitch of turns will be 2.5 cm. Now from one corner of the rectangle, draw a line on the opposite side of rectangle 2.5 cm from the opposite corner. It becomes the hypotenuse of a right triangle of sides 2.5 cm and 6 cm. 8 such lines can be drawn on the rectangular sheet which form the eight turns of the wire.

Hence, length of wire needed

= 8 * sqrt [ (2.5)^2 + (6.0)^2 ] cm

= 52 cm.

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