SOLUTIONS TO CONCEPTS CHAPTER 14 1.
F = mg
F A L Strain = L FL L F Y= AL L YA = stress = mg/A e = strain = /Y Compression L = eL F L FL y= L A L AY Lsteel = Lcu and Asteel = Acu F Stress of cu Fcu A g a) = cu 1 Stress of st A cu Fg Fst Stress =
2.
3. 4.
b) Strain = 5.
Lst FstL st A cu Ycu ( Lcu = Ist ; Acu = Ast) lcu A st Yst FcuIcu
F L L st AYst F L L cu AYcu AYcu Y strain steel wire F ( A cu A st ) cu Strain om copper wire AYst F Yst
6.
Stress in lower rod =
T1 m g g 1 w = 14 kg A1 A1
Stress in upper rod =
T2 m g m1g wg w = .18 kg 2 Au Au
For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break first. T1 m1g g 8 = 8 10 w = 14 kg A1 A1
T2 m g m1g g 8 2 = 8 10 0 = 2 kg Au Au
7. 8. 9.
The maximum load that can be put is 2 kg. Upper wire will break first if load is increased. F L Y A L F L YA L Y F L A L m2g – T = m2a …(1) and T – F = m1a …(2) m gF a= 2 m1 m2 14.1
Chapter-14 From equation (1) and (2), we get
m2 g 2(m1 m2 )
Again, T = F + m1a T
m2 g m2 g m2 g 2m1m2g m1 2 2 2(m1 m2 ) 2(m1 m2 )
Now Y =
a
FL L F A L L AY
T
m1
F
T a
L (m22 2m1m2 )g m2 g(m2 2m1 ) L 2(m1 m2 )AY 2AY(m1 m2 )
m2 m2g
10. At equilibrium T = mg When it moves to an angle , and released, the tension the T at lowest point is T = mg +
mv 2 r
The change in tension is due to centrifugal force T =
mv 2 …(1) r
Again, by work energy principle, 1 mv 2 – 0 = mgr(1 – cos) 2 2 v = 2gr (1 – cos)
…(2) m[2gr(1 cos )] 2mg(1 cos ) So, T r F = T YA L YA L F= = 2mg – 2mg cos 2mg cos = 2mg – L L YA L = cos = 1 – L(2mg) 11. From figure cos =
x
x x2 = 1 2 l l
1/ 2
x 2 l2 =x/l … (1) Increase in length L = (AC + CB) – AB 2 2 1/2 Here, AC = (l + x ) 2 2 1/2 So, L = 2(l + x ) – 100 …(2) F l …(3) Y= A l From equation (1), (2) and (3) and the freebody diagram, 2l cos = mg. FL L F 12. Y = AL L Ay
D / D D L D L L / L A 2r Again, A r 2r A r =
14.2
l
l
A T
L
Tx
L C mg
B T
Chapter-14
Pv v P = B v v
13. B =
14. 0
m m V0 Vd
d V0 0 Vd
so,
vol.strain =
V0 Vd V0
0 gh V gh 1– d = 0 (V0 Vd ) / V0 V0 B
B=
…(1)
vD 0 gh 1 v0 B
…(2)
Putting value of (2) in equation (1), we get d 1 1 d 0 (1 0 gh / B) 0 1 0 gh / B
F A Lateral displacement = l. F=Tl 2THg 4Tg 2Tg a) P b) P c) P r r r a) F = P0A b) Pressure = P0 + (2T/r) F = PA = (P0 + (2T/r)A c) P = 2T/r 2T F = PA = A r 2T cos 2T cos a) hA b) hB rA g rB g
15.
16. 17. 18.
19.
20. hHg
h
c) hC
2THg cos Hg rHgg
2T cos where, the symbols have their usual meanings. r g
Hg cos h T hHg THg cos Hg 21. h
2T cos rC g
2T cos rg
2T r P = F/r 2 23. A = r 4 3 4 3 24. R r 8 3 3 r = R/2 = 2 Increase in surface energy = TA – TA 22. P =
14.3
Chapter-14 25. h =
2T cos 2T cos , h = rg rg
hrg 2T –1 So, = cos (1/2) = 60°. 2T cos 26. a) h = rg cos =
2
b) T 2r cos = r h g
hrg 2T –3 27. T(2l) = [1 (10 ) h]g 2 28. Surface area = 4r 29. The length of small element = r d dF = T r d considering symmetric elements, dFy = 2T rd . sin [dFx = 0]
cos =
/2
so, F = 2Tr
sin d = 2Tr[cos ]
/2 0
=T2r
0
Tension 2T1 = T 2r T1 = Tr 30. a) Viscous force = 6rv
4 b) Hydrostatic force = B = r 3 g 3 4 c) 6 rv + r 3 g = mg 3
m g 2 r 2 ( )g 2 2 (4 / 3)r 3 r v= 9 9 n 31. To find the terminal velocity of rain drops, the forces acting on the drop are, 3 i) The weight (4/3) r g downward. 3 ii) Force of buoyancy (4/3) r g upward. iii) Force of viscosity 6 r v upward. Because, of air is very small, the force of buoyancy may be neglected. Thus,
4 6 r v = r 2 g 3 32. v =
or
v=
2r 2 g 9
R vD R= D
14.4
