Solved Previous Years GATE Questions on Statics
Question 1. For the loading on truss shown in the figure, the force in member CD is
(A) 0 kN
(B) 
kN
(C) 1 kN
(D) 1/
Kn
GATE-ME-2001
Hint 1. (Ans. A)
Question 2. Bodies 1 and 2 shown in the figure have equal mass m. All surfaces are smooth. The value of force P requires to prevent sliding of body 2 on body 1 is
(A) P=2mg
(B) P=
(C) 
(D) P=mg
GATE-ME-2001
Hint 2. (Ans. D)
Question.3. A truss consist of horizontal members (AC, CD, DB, and EF) and vertical members(CE and DE) having length l each. The member AE, DE and BF are inclined at 
to the horizontal. For the uniformly distributed load “P” per unit length on the member EF of the truss shown in figure given below, the force in the member CD is
(A) Pl/2
(B) Pl
(c) 0
(D) 2pl/3
GATE-ME-2003
Hint 3. (Ans. A)
Total load on EF member 
Where 
=length of the EF
For horizontal equilibrium
Taking moment about A, we have
Considering a point A.
For horizontal equilibrium
For vertical equilibrium
-ve sign shows that the force on member AC is opposite of assumed.
Now considering a point C
For horizontal equilibrium
Question 4. The figure shows a pair of pin-jointed gripper tongs holding an object weighing 2000N. The coefficient of friction (
) at the gripping is 0.1XX is the line of action of the input force and YY is the line of application of gripping force. If the pin-joint is assumed to be frictionless, then magnitude of force F required to hold the weight is
(A) 1000 N
(B) 2500 N
(c) 2000 N
(D) 5000 N
GATE-ME-2004
Hint 4. (Ans. D)
For vertical equilibrium 2Fr=2000 N
Fr=1000 N
Fr=
. RN
1000=0.1 RN
RN=1000/0.1=10000 N
Taking moment about P,
We have
0.3
F= RN
0.150
0.3F=100000.150
F=
=5000 N
Question 5. The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of
(A) 0 Newton
(B) 981 Newton in Compression
(C) 490 Newton in Compression
(D) 981 Newton in tension
GATE-ME-2004
Hint 5. (Ans A)
Consider all forces at point L
Summation of all forces along X-direction, 
Summation of all forces along Y-direction, 
Question 6. Two books of mass 1 kg each are kept on a table, one over the other. The coefficient of friction of every pair of contacting surfaces is 0.3. The lower is pulled by a horizontal force F. The minimum value of F for which slip occurs between the two books is
(A) Zero
(B) 5.74N
(C) 1.06 N
(D) 8.83 N
GATE-ME-2005
Hint 6. (Ans D)
for each surface is same =0.3
Its free body diagram (m1= m2= 1 Kg)
Equation of motion of 1
Equation of motion of 2
Now for relative motion between two blocks
Question 7. If point A is in equilibrium under the action of the applied forces, the values of tensions 
are respectively
(A) 520 N and 300 N
(B) 450 N and 150 N
(C) 300 N and 520 N
(D) 150 N and 450 N
GATE-ME-2006
Hint. 7. (Ans. A)
Solving equation (i) and (ii), we get
Question 8. If a system is in equilibrium and the position of the system depends upon many independent variables, the principle of the virtual work states that the partial derivatives of its total potential energy with respect to each of the independent variable must be
(A) -1.0
(B) 0
(C) 1.0
(D) 
GATE-ME-2006
Hint 8. (Ans. B)
Total potential energy =f (independentvariable)
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.
1. (A), 2.(D), 3. (A), 4. (D), 5. (A), 6. (D), 7. (A), 8. (B)
One Response to “Previous Years GATE MCQ’s on Statics 2001-2006”
Akshata Jaiswal
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