RESULTANT & EQUILIBRIUM Example problems regarding resultant and equilibrium.
ESQUIVEL, J.
R ESULTANT
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1
QUESTION no. : A boat is headed due east at a forward speed of 2 knots (kn). Then it headed northeast at a speed of 4kn. Then it made a turn of southwest of 6kn. Disregarding other pressures, what is the velocity of the boat relative to the earth’s surface? 1kn = 1.852 km/h = 1.151 mi/h
SOLUTION: Ax = 2
Bx = 4cos45º
Ay = 0
= 2.83
= -2.83
= -4.24
By = 4sin45º
Cy = 4sin45º
Dy = 6sin225º
= 2.83
= 2.83
= -4.24
Rx = Ax + Bx + Cx + Dx = 2 + 2.83 + (-2.83) + (-4.24)
Cx = 4cos135º
Dx = 6cos225º
R = 2.6kn.
Rx = -2.24 tanө = Ry = Ay + By + Cy + Dy = 0 + 2.83 + 2.83 + (-4.24) Ry = 1.42
= = 0.63 tan-1 Ө = 32.19º
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QUESTION no.
2:
A scout begins his trip in the woods by first walking 4km southeast from his base camp. Then he walks 6km in the direction 60º north of east. Determine the magnitude of the resultant for the trip.
SOLUTION:
R2 = A2 + B2 – 2AB cosө = 42 + 62 – 2(4)(4) cos(180º-45º-60º) = 16 + 36 – 48cos75º = 52 – 48(0.2588) = 52 – 124.42 = R = 6.3km
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3
QUESTION no. :
The component of Vector B is along the positive x-direction which measures 8 units, and its component in the negative y-direction is 12 units. Determine the magnitude of Vector B and angle ß that it makes with the positive xdirection.
SOLUTION: Bx = 8 and By = 12
B= = = = 14.42
ß = tan-1 = tan-1 = 56.31º
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4
QUESTION no. : A boy scout walks 22 km in North direction, and then he walks in direction 60º Southeast during 47.0 km. Find the components of the resulting vector displacement from the starting point, its magnitude and angle with the x axis.
SOLUTION:
D2 = Dx2 + Dy2 = (23.5) 2 + (-18.7)2 = 552.25 + 349.69
tanß = = = -0.80 tan-1
= 901.94 ß = 38.5 SouthEast D= 30.03
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5
QUESTION no. :
I move 3 meters at 25 degrees North of East and then 6 meters 40 degrees West of North. How far from the starting point I have moved?
SOLUTION:
= Ax + Ay
= Bx + By
= (3cos25º)x + (3sin25º)y
= (6sin40º)x + (6cos40º)y
= (2.72)x + (1.27)y
= (-3.86)x + (4.60)y
+
= (2.72-3.86)x + (1.27 + 4.60)y = (-1.14)x + (5.87)y
+
= = 5.98m
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6
QUESTION no. : The angle of elevation to the top C of a building from two points A and B on level ground are 50 degrees and 60 degrees respectively. The distance between points A and B is 30 meters. Points A, B and C are in the same vertical plane.
SOLUTION: B = 180º - 60º
Sine law:
= 120º
=
C = 180º - (50º + 120º)
d=
= 10º = 135.18
sin60 = h = dsin60 = 135.18sin60 h = 117.07
tan50 = = 1.19 =
35.7 = 0.87 AC – 0.595 AC 129.82 = AC
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7
QUESTION no. :
Ax = 28cos15º = 27.05N
Bx = 15cos75º = 3.88N
Ay= 28sin15º
By= 15sin75º
= 7.25N
= 14.49N
Cx = 0 Cy = -11N
SOLUTION: ∑x = Ax + Bx + Cx = 27.05 + 3.88 + 0 ∑x = 30.93
∑y = Ay + By + Cy
R= =
= = 32.74
Өx = =
tan-1 tan-1
= 2.78tan-1 x = 70.85
= 7.25 + 14.49 + (-11) ∑y = 10.74
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8
QUESTION no. :
Fax = 0 Fay= 150 Fbx = 125 Fby= 0 Fcx = -125cos45º = -88.39 Fcy = -125sin45º = -88.39
SOLUTION: ∑x = Fax + Fbx + Fcx = 0 + 125 + (-88.39) ∑x = 36.61
∑y = Fay + Fby + Fcy
tan-1
R =
Өx =
=
=
=
= 1.68tan-1
=
=
tan-1
59.28
= 71.67
= 150 + 0 + (-88.39) ∑y = 61.61
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9
QUESTION no. : Ax = -3.4cos19º = -3.21 Ay = 3.4sin19º = 1.11 Bx = 9.2cos70º = 3.15 By = 9.2sin70º = 8.65
SOLUTION: ∑x = Ax + Bx = -3.21 + 3.15 ∑x = 0.06
R=
Өx =
=
=
= =
∑y = Ay + By + Fcy
tan-1
= 162.67tan-1 =
= 9.76
tan-1
89.65
= 1.11 + 8.65 ∑y = 9.76
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10: :
QUESTION no.
SOLUTION: ∑x = 20cos60º = 10
∑y = 20sin60º
tan-1
R =
Өx =
=
=
=
= 1.732tan-1
R = 19.99
=
tan-1
59.99
= 17.32
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11:
QUESTION no.
Ax = 86 Ay= 0
Bx = 67cos40º = 51.32
∑x = Ax + Bx + Cx = 86 + 51.32 + (-13) ∑x = 124.32
= -13
By = 67sin40º
Cy = -26sin60º
= 43.07
= -22.52
tan-1
R =
Өx =
=
=
=
= 0.16tan-1
= 126.01 ∑y = Ay + By + Cy
Cx = -26cos60º
x = 9.39
= 0 + 43.07 + (-22.52) ∑y = 20.55
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tan-1
12:
QUESTION no.
h =
tan Ө =
=
=
=
= 1 tan-1
h = 28miles
Ө = 45
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13:
QUESTION no.
To solve for side b, use Sine Law: = =
R2 = a2 + b2 = (15m)2 + (30m)2 = 225 + 900 =
30 ºb = 900 R = 33.54 = b = 30
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14:
QUESTION no.
A Philippine Airlines leaves Ninoy Aquino International Airline on an overcast day and is later sighted 275km away, in a direction making an angle of 30ºeast of north. How far east and north is the airplane from the airport when sighted?
SOLUTION: Let dx and dy be the horizontal and vertical displacement = cosө = cos60º dx = 275(0.5) = 137.5
= sinө = sin60º dy = 275(0.87) dy = 238.25
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15:
QUESTION no.
SOLUTION: R= = = =
16:
QUESTION no.
SOLUTION: R2= a2 + b2 = 900 + 1600
tanө = = 1.33 tan-1 Ө = 53.13
= R = 50
17:
QUESTION no.
SOLUTION: R2 = a2 + b2 = 25 + 100
tanө = = 0.5 tan-1 Ө = 26.57
= R = 11.18 Page | 17
E Q U I L I B R I UM
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1
QUESTION no. :
An object has a weight of 125lbf. The object is suspended by cables. Calculate the tension (T1) in the cable 30º with the horizontal.
FREE BODY DIAGRAM:
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SOLUTION: ∑F = T1x + T2x + T3x = 0 ∑F
= T1x + T2x + T3x = 0
(T1 sin30º) + (T2sin180º) + (T3sin270º) = 0 (0.5 T1) + (0T2) + [(125lbf) (-1)] = 0 0.5 T1 = 125lbf T1 = 250lbf
2
QUESTION no. :
A rope extends between two poles. A 90-N boy hangs from it as shown in the figure. Find the tensions in the two parts of the rope.
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FREE BODY DIAGRAM:
SOLUTION:
∑F = 0
∑F
FT2cos5.0º - FT1cos10º = 0
FT2sin5.0º + FT1sin10º - 90N = 0
0.966 FT2 – 0.985 FT1 = 0
0.087FT2 + 0.174 FT1 - 90N = 0
FT2 = 0.99 FT1
0.087(0.99FT1) + 0.174FT1 = 90N
FT2 = 0.99 (0.35kN)
0.086FT1 + 0.174FT1 = 90N
FT2 = 0.34kN
FT1 = 0.35kN
=0
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3
QUESTION no. :
The tension in the horizontal cord is 30N. Find the tension on cord 2 and the weight of the object.
FREE BODY DIAGRAM:
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SOLUTION:
∑F
∑F
=0
=0
30N – FT2cos40º = 0
FT2sin40º – Fw = 0
Since FT1 = FW,
FT2 (0.77) = 30
Fw = 38.96 (0.64)
FT1 = 25.04
FT2 = 38.96
= 25.04
4
QUESTION no. :
Find the tensions in the ropes if the supported object weighs 600N.
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FREE BODY DIAGRAM:
SOLUTION: ∑F = 0
∑F
FT2cos60º - FT1cos60º = 0
FT1sin60º + FT2sin60º - 600N = 0
=0
0.5 FT2 – 0.5 FT1 = 0 0.87FT1 + 0.87 FT1 - 600N = 0 FT2 = FT1 1.74FT1 = 600N FT2 = 344.8 FT1 = 344.83
∑F = 0
∑F
FT3cos20º - FT5 - 346sin30º = 0
FT3sin20º - 346cos30º = 0
FT3 = 877N
FT5 = 651N
=0
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5
QUESTION no. :
The uniform 0.60kN beam is hinged at P. Find the tension in the tie rope and the components of the reaction force exerted by the hinge on the beam.
FREE BODY DIAGRAM:
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SOLUTION:
Torque equation on P, + (3L/4)(T)(sin40º) – (L)(800N)(sin90º) – (L/2)(600N)(sin90º) = 0 FT = 2280N
∑F = 0
∑F
-FTcos40º + FRH = 0
FTsin40º + FRV – 600N – 800= 0
FRH = 1750N or 1.8kN
FRV = 65.6N or 66N
=0
6
QUESTION no. :
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FREE BODY DIAGRAM:
SOLUTION: ∑F = T1cos70º - T2 cos19º = 0
∑F
T1 =
(2.78 T2) sin70 º + T2sin19º = 9.8N
T1 = 2.78 T2 T1 = 2.78 (3.33N)
= T1sin70º
+ T2sin19º - 9.8N = 0
2.61 T2 + 0.33 T2 = 9.8N T2 = 3.33N
T 1 = 9 .2 6 N
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7
QUESTION no. :
The object weighs 50N and is supported by a cord. Find the tension in the cord.
FREE BODY DIAGRAM:
SOLUTION: ∑F = 0
∑F
0=0
FT – 50N = 0
=0
FT = 50N
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8
QUESTION no. :
A 5.0 kg object is to be given an upward acceleration of 0.30m/s2 by a rope pulling straight upward on it. What must be the tension in the rope?
FREE BODY DIAGRAM:
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SOLUTION: Fw = mg = (50kg)(9.81m/s2) = 49.1N.
∑F
= may
FT – mg = may FT – 49.1N = (5.0kg) (0.30m/s2) FT = 50.6N
9
QUESTION no. : You hang your picture frame by means of vertical string. Two strings in turn support this string. Each string makes 30º with an overhead horizontal beam. Find the tension in the strings.
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FREE BODY DIAGRAM:
SOLUTION:
∑F = 0
∑F
T1cos30º - T2 cos30º
T1sin30º + T2sin30º - 55N = 0
T1cos =T2
Since T1 = T2, we may replace T2 by T1 and solve the above equation for T1 .
∑F
=0
=0
T1 (0.5) + T1 (0.5) – 55N = 0 T1 = 5 5 N T2 = 5 5 N
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10:
QUESTION no.
A 2000lbm car is accelerating (on a frictionless surface) at a rate of 2ft-sec. What force must be applied to the car to act as an equilibrant for this system?
FREE BODY DIAGRAM:
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SOLUTION:
A Force, F2, must be applied in the opposite direction to F1 such that the sum of all forces acting in the car is zero. ∑ Forces = F1 + F2 + N + W = 0
Since the car remains on the surface, forces N and W are in equal and opposite directions. Force F2 must be applied in an equal and opposite direction to F1 in order for the forces to be in equilibrium. F2 = F1 =
= (2000lbm x 2ft-sec2) / 32.17
= 124lbf
11:
QUESTION no.
A block of weight w hangs from a cord which is knotted at 0 to two other cords fastened to the cords. Let w=50lb, Ө2=30º, and Ө3=60º. The weights of the cords are negligible.
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FREE BODY DIAGRAM:
(a)
(b)
(b)
(a) Forces acting on the block, on the knot, and on the ceiling. (b) Forces on the knot 0 resolved into x- and y- components.
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SOLUTION: Tì = w = 50lb Since T1 and Tì form an action-repair pair, Tì = T1 Hence T1 = 50lb.
To find the forces T2 and T3, we resolve these forces into rectangular components. Then, from Newton’s second law, ∑F = T2cosӨ2 – T3 cosӨ3 = 0 ∑F
= T2sinӨ2
We have ∑F ∑F
– T3 sin Ө3 - T1 = 0
= T2cos30º – T3 cos60º = 0 = T2sin30º – T3 sin 60º - T3 = 50
Or ∑F = 0.866 T2 – 0.500 T3 = 0 ∑F
= 0.500 T2 + 0.866 T3 = 0
Solving these equations simultaneously, we find the tensions to be T2 = 25lb
T3 = 43.3 lb.
We know from Newton’s third law that the inclined cords exert on the ceiling the forces T 2 and T3, equal and opposite to T2 and T3, respectively.
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12:
QUESTION no.
A 10-kg sign is suspended by two ropes, each supporting an equal portion of the sign's weight. The two end ropes make an angle of 30º and 45º to the horizontal, respectively. What is the tension on each of the ropes?
FREE BODY DIAGRAM:
SOLUTION: ∑F = Facos30º - Fbcos45º = 0 Fa = Fa = 0.81 Fb Fa = 0.81 (8.97)
∑F
= Fasin30º
+ Fbsin45º - 10 = 0
(0.81Fb)sin30º + Fbsin45º = 10 0.405Fb + 0.71 Fb = 10 Fb = 8.97
= 7.26
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13:
QUESTION no.
The weights of the objects are 200N and 300N. The pulleys are essentially frictionless and massless. Pulley P1 has a stationary axle, but pulley P2 is free to move up and down. Find the tensions FT1 and FT2 and the acceleration of each body.
SOLUTION: FT1 = 2FT2 Let a = the downward acceleration of A. a/2 = upward acceleration of B.
∑Fy = may FT1 – 300N = (mB)( a) ; 200N - FT2 = mAa But m = mA = (200/9.81)kg mB = (300/9.81)kg
By substituting these equations, we can compute FT1, 2FT2, and a. FT1 = 327N
FT2 = 164N a = 1.78 Page | 37
14:
QUESTION no.
A uniform, 0.40-kN boom is supported as shown in the figure. Find the tension in the tie rope and the force exerted on the boom by the pin at P.
FREE BODY DIAGRAM:
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SOLUTION:
Torque Equation: + (3L/4)(FT)(sin50º) – (L/2)(400N)(sin40º) – (L)(2000N)(sin40º) FT = 2460N or 2.5kN.
∑F = 0
∑F
FRH – FT = 0
FRV – 2000N – 400N = 0
FRH = 2.4 kN
FRV = 2.4 kN
=0
Since FRH and FRV are the components of the reaction force at P
= 3.4kN
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15:
QUESTION no.
A 200lb man hangs from the middle of a tightly stretched rope so that the angle between the rope and the horizontal direction is 5º , as shown in the Figure. Calculate the tension in the rope.
FREE BODY DIAGRAM:
SOLUTION: ∑F = 0
∑F
T1cos30º - T2 cos30º = 0
T1 sin5º + T2sin5º - 200 lb = 0
T1 = T2 T1 = 555.56
=0
200lb = 0.09T2 + 0.09T2 = T2 555.56 = T2
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