11. An electric pole is supported by a wire which exerts a pull of 700 N on the top of the pole. If the angle between the wire and the pole is 195° with + ve x– axis. What are the horizontal and vertical components?
(a) 181.17 N, 676.14 N (b) 676.14 N, 181.17 N (c) 857.31 N,
181.17 N (d) 181.17, 857.31 N
Ans.: (a) Explanation : F = 700 N q = 270°– 195 = 75° SFx
= F× cos q = 700 ´ cos 75°= 181.17 N SFy = F× sin q = 700 ´ sin 75°= 676.14 N
12. If a wooden block of weight W = 97 N placed on a rough surface inclined at an angle q = 35° in anticlockwise direction with positive x– axis and is at rest, then assuming up the plane forces positive, the reaction offered by the surface is
(a) 97 N (b) 56 N (c) 80 N (d) 0 N
Ans.: (c) Explanation : Normal Reaction = F sin q = 37 ´ sin 55 = 79.45 » 80 N
13. Four forces 18 N, 36 N, 54 N and 72 N are acting along sides AB, BC, CD and DA of a rectangle ABCD of side (2´ 3) m. Their resultant forces is 150 N. Calculate position of resultant w.r.t. ‘A’
(a) 1.56 m (b) 15.60 m (c) 156 m (d) 0.156 m
Ans.: (a) Explanation : Use Varignon¢s theorem R × x = (– 36 ´ 2) – (54 ´ 3) 150 ´ x = – 72 – 162 150 = –
234 x = 234 150 = 1.56 m
14. Two like parallel forces 38 N and 86 N are acting at a distance of 6cm. What is the Resultant and Position?
(a) 416 N, 1.24 cm from Q (b) 124 N, 4.16 cm from Q (c) 124
N, 4.16 cm from P (d) 416 N, 1.24 cm from P
Ans.: (c) Explanation : R = 38 + 86 = 124 N Use Varignon¢s theorem R × x = SMp 124 x = 86 ´ 6 = 516 x = 4.16 cm
15. A force of magnitude 125 N is directed from point A (2, 3) to B (5, 9). The x and y components are
(a) –100 N, 75 N (b) 100 N, 75 N (c) 75 N, 100 N (d) 100 N,
– 75 N
Ans.: (c) Explanation : tanq
= y2– y1 x2– x1 = 7 – 3 5 – 2 = 4 3 = 1.34 q
= 53.26°, F =
125 N Fx = Fcos q, Fx
= 125 cos q= 75
N Fy = F sin q, Fy
= 125 sin q= 100
N
16. A force of magnitude 125 N is directed from A(2, 3) to B
(– 5, –7). The x and y components are
(a) –83.05 N, 93.41 N (b) – 93.41 N, 83.05 N (c) 83.05 N,
93.41N (d) 93.41 N, 83.05 N
Ans.: (c) Explanation : tanq
= y2– y1 x2– x1 = – 7 – 2 – 5 – 3 = – 9 – 8 = 1.125 q = 48.36° Fx = F cos q = 125 ´cos 48.36°= 83.05 N Fy = F sin q = 125´ sin 48.36°= 93.41 N
17. A force of 20 N is acting along a line having slope 3 4
. Calculate Resultant.
(a) 16 N (b) 20 N (c) 12 N (d) 28 N
Ans.: (b) Explanation : Slope tan q = 3 4 q = 36.86° Fx = f cos q = 20 cos 36.86° = 16 Fy = f sin q = 20 sin 36.86° = 12 R = 20 N
18. The moment of a force about any point is geometrically equal to ____ area of triangle where base represents the force and height represents the perpendicular distance.
(a) Half (b) Twice (c) Same (d) thrice
Ans.: (b) Explanation : MeC = F ´ d = Fd A (DACB) = 1 2 AB ´ CD = 1 2 ´ F ´ d = 1 2 ´ F d i.e = 1 2 Mc i.e M e c = 2 A
(D ABC)
19. If the arm of couple is doubled, its moment will
(a) Be halved (b) Remains same (c) Be doubled (d) None of the
above
Ans.: (c) Explanation
: Case – I) M = 10 ´2 =
20 Nm Case – II) M = 10 ´4
= 40 Nm
20. A force of 200 N acting download tangentially to a drum of radius 0.75 m, must be transferred parallel to itself to its centre 0. The moment which should accompany it for equivalent effect is
(a) 150 Nm (b) – 150 Nm (c) – 75 Nm (d) 75 N.m
Ans.: (b) Explanation : M = f×
d M0 = 200 ´ 0.75
= 150 Nm \
Equivalent effect = 150 NmM
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