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# Machine Kinematics MCQ – Velocities in a Slider Crank Mechanism and Motion of a Link

The lengths of the links of a 4- bar linkage with revolute pairs only are p,q,r and s units. Given that p < q < r < s and s+p < q+r which of these links should be the fixed one, for obtaining a ‘double crank’ mechanism?

a) ink of length p

b) link of length q

c) link of length r

d) link of length s

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Answer: a

Explanation: For Double crank mechanism Shortest link is fixed.

Here shortest link is ‘P’.

For a four-bar linkage in toggle position, the value of mechanical advantage is

a) 0.0

b) 0.5

c) 1.0

d) ∞

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Answer: d

Explanation: At Toggle position output velocity is zero

And hence, mechanical advantage = input velocity/output velocity = ∞.

In a slider-crank mechanism, the crank is rotating with an angular velocity of 20 rad/s in counterclockwise direction. At the instant when the crank is perpendicular to the direction of the piston movement, velocity of the piston is 2 m/s. Radius of the crank is

a) 100 cm

b) 10 cm

c) 1 cm

d) 0.1 cm

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Answer: b

Explanation: Vp = ωr(sinϴ + sin2ϴ/2n)

In this case ϴ = 900

Vp = ωr

r = Vp/ω = 2/20 = 0.1 m or 10 cm.

In a single link robotic arm the end-effector slides upward along the link with a velocity of 0.5 m/s while the link rotates about revolute joint with an angular speed of 1 rad/sec. When the end-effector is at a distance of 1 m from the joint, the acceleration experienced by the end-effector will be

a) 1 m/s2

b) 1.41 m/s2

c) 1.71 m/2

d) 2 m/2

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Answer: a

Explanation: a = 2ωV = 2 x 1 x 0.5 = 1 m/s2.

For the same crank length and uniform angular velocity of the crank in an offset slider crank mechanism, if the connecting rod length is increased by 1.5 times, the velocity of piston will

a) remain unchanged

b) increase 1.5 times

c) decrease by 1.5 times

d) increase by 1.5√2 times

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Answer: c

Explanation: V1 = ωr(sinϴ + sin2ϴ/2n)

V2 = ωr(sinϴ + sin2ϴ/3n)

from these two equation, V2 < V1

V2 will decrease but correct quantification can not be done with available data.

Among the available options, best answer is (c).

It is planned to construct a four-bar mechanism ABCD with length AB= 60mm, BC = 100mm, CD = 70 mm and fixed link AD = 200 mm. If at least one link is required to have a complete rotation, this mechanism is

a) of crank-rocker type

b) of double-crank type

c) of double rocker type

d) impossible to construct

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Answer: c

Explanation: S + L = 60 + 200 = 260 mm

P + Q = 100 + 70 = 170 mm

From grashoff equality when S + L > P + Q

So always double rocker.

The number of links in a planer mechanism with revolute joints having 10 instantaneous centres is

a) 3

b) 4

c) 5

d) 6

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Answer: c

Explanation: n(n – 1)/2 = 10

n(n – 1) = 20

n = 5.

A weston differential pulley block consists of a lower block and upper block. The upper block has two cogged grooves, one of which has a radius of 150 mm and the other a radius of 125 mm. If the efficiency of the machine is 50% calculate the effort required to raise a load of 1.5 kN.

a) 250 N

b) 300 N

c) 350 N

d) 400 N

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Answer: a

Explanation: We know that in case of a Weston differential pulley block,

V.R. = 2D/D – d = 2 x 300/300 -250 = 12

Using the relation, Efficiency = M.A./V.R. x 100

or, 50 = M.A./12 x 100

M.A. = 6

Again, M.A. = W/P

6 = 1.5 x 1000/ P

P = 250 N.

Following are the specifications of a single purchase crab:

Diameter of load drum, d = 200 mm

Length of lever, l = 1.2 m

No. of teeth on pinion, T1 = 10

No. of teeth on spur wheel, T2 = 100

Find the velocity ratio of this machine.

a) 100

b) 110

c) 120

d) 130

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Answer: c

Explanation: V.R. = 2l/d x T2/T1

= 2 x 120/20 x 100/10 = 120.

On a machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the law of the machine.

a) P = 1/100W + 60

b) P = 1/100W + 70

c) P = 1/100W + 80

d) P = 1/100W + 90

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Answer: c

Explanation: Let the law of machine be P = mW + C

where P = effort applied, W = load lifted and m and C being constants.

when P = 100 N W = 3000 N

when P = 160 N W = 9000 N

Putting these values in the law of machine.

100 = 3000m + C …………(i)

160 = 9000m + C …………(ii)

Subtracting (i) and (ii), we get

60 = 6000 m

or, m = 1/100

Putting this value in equation (i), we get

100 = 3000 x 1/100 + C

C = 70

Hence, the machine follows the laws

P = 1/100W +70.

Following are the specifications of a single purchase crab:

Diameter of load drum, d = 200 mm

Length of lever, l = 1.2 m

No. of teeth on pinion, T1 = 10

No. of teeth on spur wheel, T2 = 100

On this machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the efficiency at 3000N.

a) 10 %

b) 15 %

c) 20 %

d) 25 %

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Answer: d

Explanation: V.R. = 2l/d x T2/T1

= 2 x 120/20 x 100/10 = 120

M.A. = W/P = 3000/100 = 30

Efficiency = M.A./V.R. = 30/120 = 0.25 = 25%.

Following are the specifications of a single purchase crab:

Diameter of load drum, d = 200 mm

Length of lever, l = 1.2 m

No. of teeth on pinion, T1 = 10

No. of teeth on spur wheel, T2 = 100

On this machine efforts of 100 N and 160 N are required to lift the loads of 3000 N and 9000 N respectively. Find the efficiency at 9000N.

a) 30 %

b) 40 %

c) 46.8 %

d) 56.8 %

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Answer: d

Explanation: V.R. = 2l/d x T2/T1

= 2 x 120/20 x 100/10 = 120

M.A. = W/P = 9000/160 = 900/16

Efficiency = M.A./V.R. = 900/16 x 120 = 0.468 = 46.8%.