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Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Mechanical Engineering MCQ


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20 Questions MCQ Test - Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work

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Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 1

A spring scale indicates a tension T in the right hand cable of the pulley system shown in Neglecting the mass of the pulleys and ignoring friction between the cable and pulley the mass m is

[1995]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 1


∑ FV = 0
4T = mg
or m = 4T/g

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 2

A rigid ball of weight 100 N is suspended with the help of a string. The ball is pulled by a horizontal force F such that the string makes an angle of 30° with the vertical. The magnitude of force F(in N) is ______.

[2016]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 2


Applying Lami’s theorem

F = 57.74 N

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Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 3

A force F is acting on a bent bar which is clamped at one end as shown in the figure.

The CORRECT free body diagram is

[2016]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 3

Fixed support should be replaced by vertical reaction, horizontal reaction and a moment reaction in free body diagram.

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 4

A bar of uniform cross section and weighing 100 N is held horizontally using two massless and inextensible strings S1 and S2 as shown in the figure.

[2018]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 4


Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 5

If point A is in equilibrium under the action of the applied forces, the values of tensions TAB and TAC are respectively.

[2006]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 5


TAB cos 60º = TAC cos 30º ...(i)
TAB sin 60º + TAC sin 30º = 600 ...(ii)
Solving equations (i) and (ii), we get
TAB = 520N and TAC = 300N

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 6

A weight of 500 N is supported by two metallic ropes as shown in the figure. The values of tensions T1 and T2 are respectively

[2015]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 6

Using sine rule


T1 = 500 x sin 120º
T2 = 500 sin 150º
T1 = 433 N, T2 = 250N

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 7

A car having weight W is moving in the direction as shown in the figure. The center of gravity (CG) of the car is located at height h from the ground, midway between the front and rear wheels. The distance between the front and rear wheels is l. The acceleration of the car is a, and acceleration due to gravity is g. The reaction on the front wheels(Rf) and rear wheels (Rr) are given by:

[2019]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 7

8. Here, Reaction on front wheels = Rf Reaction on rear wheels = Rr

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 8

A mass 35 kg is suspended from a weightless bar AB which is supported by a cable CB and a pin at A as shown in figure. The pin reactions at A on the bar AB are

[1997]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 8


For block, T2 = mg = 343.35 N ...(i)

∴ T1 = 829.74 N
∑Fy = O
Ry – T2 + T1 sinθ = O
∴ Ry = O
∑Fx = O
Rx – T1 cosθ = O
Rx = 755.4 N

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 9

A truss consists of horizontal members (AC, CD, DB  and EF) and vertical members (CE and DF) having length l each. The-members AE, DE and BF are inclined at 45° to the horizontal For the uniformly distributed load p per unit length on the members EF of the truss shown in figure given below, the force in the member CD is

[2003]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 9

Total load on EF member = pl
where l = length of EF
For horizontal equilibrium
RA + RB = Pl

Taking moment about A, we have

Considering a point A.
For horizontal equilibrium FAC + FAE cos 45º = 0

For vertical equilibrium
RA – FAE sin 45º = 0


–ve sign shows that the force on member AC is opposite of assumed. Now considering a point C For horizontal equilibrium

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 10

The figure shows a pin-jointed plane truss Ioaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of

[2004]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 10


Consider all forces at point L

Summation of all forces along x-direction,
FLK = FLM
Summation of all forces along y-direction,
FLN = 0

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 11

Consider a truss PQR loaded at P with aforce Fas shown in the figure.

The tension in the number QR is

[2008]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 11

TPQ cos 40° + TPR cos 60° + F = 0
and  TPQ sin 40° = TPR sin 60°
From these equations
TPQ and TPR can be found in terms of F

Now at Q, T TQR = TPQ cos 45°
Solving these equations, we get
TQR = 0.63 F

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 12

Two steel truss members, AC md BC, each having cross sectional are of 100 mm2, are subjected to a horizontal force F as shown in figure. All the joints are hinged.

If F = 1 kN, the magnitude of the vertical reaction force developed at the point B in kN is

[2012]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 12

Vertical reaction at point B,
RB = FB cos 30 = 0.633 F
F = 1k N
∴ RB = 0.633 kN

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 13

Two steel truss members, AC md BC, each having cross sectional are of 100 mm2, are subjected to a horizontal force F as shown in figure. All the joints are hinged.

The maximum force F in kN that can be applied at C such that the axial stress in any of the truss members DOES NOT exceed 100 MPa is

[2012]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 13

From Lami’s theorem,

⇒ FA = 0.8965 F; FB = 0.732 F
Hence maximum force is FA
and stress = FA/Area = 100 MPa
∴ 0.8965 F/100  = 100 MPa
⇒ F = 11154.48 N   or 11.15 k N

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 14

For the truss shown in the figure, the forces F1 and F2 are 9 kN and 3 kN, respectively. The force (in kN) in the member QS is (All dimensions are in m)

[2014]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 14

By method of sections

∑V = 0
F1 = FQS si n 53.1
⇒ FQS = 11.25(T)

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 15

For the truss shown in figure, the magnitude of the force in member PR and the support reaction at R are respectively

[2015]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 15

∑MQ = 0
100 × cos 60 × 4 = Ra × 4
⇒ Ra = 50 kN
FPR × cos 45 = 100 cos 60
⇒ FPR = 70.71 kN

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 16

For the truss shown in the figure, the magnitude of the force (in kN) in the member SR is

[2015]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 16


Rp + Rx = 30000
∑MP = 0
Rx x 3 = 2 x 3000
⇒ Rx = 20 kN and Rp = 10 kN

For balance at ‘x' → FRx = 20 kN
at ‘RT’ → FRT cos 45 = 20
 (i)
Also FSR = FRT cos 45 ... (ii)
From equations (i) and (ii),
FSR = 20kN

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 17

A two member truss PQR is supporting a load W. The axial forces in members PQ and QR are respectively

[2016]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 17


Lami’s theorem is used to find the F

⇒ TPQ = √3W(T)
⇒ TQR = – 2W = 2W(C)

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 18

If a system is in equilibrium and the position of the system depends upon many independent variables, the principle of virtual- work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be

[2006]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 18

Total potential energy = f (independent variable) Hence for a system in equilibrium, Total potential energy = Constant Thus, partial derivatives of its total potential energy with respect to each of independent variable must be zero.

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 19

Two identical trusses support a load of 100 N as shown in the figure. The length of each truss is 1.0 m, cross-sectional area is 200 mm2; Young's modulus E = 200 GPa. The force in the truss AB (in N) is

[2015]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 19

⇒ 2F sin 30 = 100
⇒ F = 100 N

Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 20

A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in newton) required to maintain equilibrium of the ladder is_____.

[2014]

Detailed Solution for Past Year Questions: FBD, Equilibrium, Plane Trusses And Virtual Work - Question 20

Taking moments about B
∑MB = 0
⇒ W × 2 = RA × 3

∑H = 0
⇒ RA = P = 400 N

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