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JNTU B.Tech I Year ENGINEERING MECHANICS Previous Question Paper (July - 2021 - R16)

B.Tech | I Year | I-I | R16 | July - 2021

Answer PaperPaper Code: 131AEMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

Figure 1 shows the body under the action of coplanar forces. Determine the magnitude, direction and position of the single force which will keep the body in equilibrium.

15 marks

Answer

To find the single force (equilibrant) that keeps a body in equilibrium under a system of coplanar forces, we must determine the resultant force of the system and then apply an equal and opposite force at the correct position.

The steps to solve this are:

  1. Resolve all given forces into their horizontal (ΣFx) and vertical (ΣFy) components.
  2. Calculate the magnitude of the resultant force (R) using the formula:
    R = √(ΣFx2 + ΣFy2)
  3. Determine the angle (θ) of the resultant force with respect to the x-axis:
    θ = tan-1(|ΣFy / ΣFx|)
  4. Calculate the sum of moments (ΣMO) of all forces about an arbitrary point (usually the origin) to find the position of the resultant.
  5. Determine the perpendicular distance (d) of the resultant from the reference point using:
    d = |ΣMO / R|
  6. The equilibrant force (E) has the same magnitude (R) and the same line of action as the resultant, but acts in the opposite direction (180° + θ).

By applying this equilibrant force at the calculated distance from the reference point, the net force and net moment acting on the body become zero, satisfying the conditions for static equilibrium.

2

Two persons lift a man of 100 kg by cables passing over two pulleys as shown in figure 2. Determine the P and Q that must be applied by two persons if the body is in equilibrium at the position shown.

15 marks

Answer

To determine the forces P and Q required to maintain the man of mass 100 kg in equilibrium, we apply the conditions of static equilibrium. First, calculate the weight of the man (W):

W = m × g = 100 kg × 9.81 m/s² = 981 N

Assuming the cables are attached to the man at a common point, the system is in equilibrium when the sum of vertical and horizontal forces equals zero. Let the angles of the cables with the vertical be θ₁ and θ₂ respectively. Resolving the forces into components:

  1. Sum of horizontal forces: P sin(θ₁) - Q sin(θ₂) = 0, which implies P sin(θ₁) = Q sin(θ₂)
  2. Sum of vertical forces: P cos(θ₁) + Q cos(θ₂) = W = 981 N

By solving this system of two equations, the tension in each cable is determined. If the angles are provided in the figure (e.g., θ₁ = 30° and θ₂ = 45°), substitute these values into the equations:

P sin(30°) = Q sin(45°) ⇒ P(0.5) = Q(0.707) ⇒ P = 1.414Q
P cos(30°) + Q cos(45°) = 981

Substituting P = 1.414Q into the vertical equilibrium equation:

(1.414Q)(0.866) + Q(0.707) = 981
1.225Q + 0.707Q = 981
1.932Q = 981 ⇒ Q = 507.76 N
P = 1.414 × 507.76 = 717.97 N

The forces P and Q depend directly on the specific geometry (angles) shown in the figure; using the equilibrium equations above allows for the precise calculation once those angles are identified.

3

Determine the least value of P to cause motion to impend rightwards shown in figure 3. Assume the pulley to be frictionless and coefficient of friction of all contiguous surfaces is 0.15.

15 marks

Answer

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4

Block A supports a weight of 4000 N and it is to be prevented from sliding down by applying a horizontal force P on the block B, shown in figure 4. If the coefficient of friction at all surfaces of contact 0.2, determine the smallest force P required to maintain equilibrium. Neglect the weight of the blocks.

15 marks

Answer

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5

Determine the centroid of the shaded area ABDEFG shown in figure 5.

15 marks

Answer

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6

Find the moments of inertia of the shaded area, shown in figure 6, about its centroidal axes.

15 marks

Answer

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715 marks
a

Derive the transfer formula for mass moment of inertia.

8 marks

Answer

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b

Find the mass moment of inertia of a solid cone about its centroidal axis. Assume data required.

7 marks

Answer

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8

The two step pulley in the figure 7 has weight 1800 N and radius of gyrations k = 175 mm. Develop a formula for the downward acceleration of the falling weight on the right if W = 220 N, r1= 250mm and r2 = 350mm.

15 marks

Answer

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This page contains the question paper of ENGINEERING MECHANICS from JNTU for B.Tech PTM I Year I-I, July - 2021, R16.

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