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For the materials handling mechanism described in Problem, determine the torque required from the motor if...

Question

For the materials handling mechanism described in Problem, determine the torque required from the motor if...

For the materials handling mechanism described in Problem, determine the torque required from the motor if the motor is rotat

For the materials handling mechanism described in Problem, determine the torque required from the motor if the motor is rotating at 120 rpm and decelerating at a rate of 100 rad/s2. The materials handling mechanism, shown in Figure, slides 4 kg packages along a counter. The machine operates with the crank rotating counterclockwise at a constant rate of 120 rpm. The coefficient of kinetic friction between the package and counter is 0.15. The weight of all the mechanism links is negligible. Determine the instantaneous torque required from the motor to operate this mechanism. FIGURE 0.65 m 0.32 m 0.4 m 0.4 m 0.1 m 30° 0.2 m 0.6 m 0.3 m 0.37 m
Engineering Mechanical Engineering
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Answers

Solution:

The simplified kinematic diagram:

065 5 E 0.32 0.3 m B 3 C. 02 06m 03 m D

Calculate the angular velocity of crank:

= لي 27TV 60

2 XT X 120 60

= 12.56 rad S

Calculate the velocity of the point B:

VB =TAB X W

= 0.1 x 12.56

1.256 m/s

As the angular velocity is directed perpendicular to the movement of link, the orientation of VB is 60° north of west.

Velocity of link C:

= VB + VC/B

Velocity of link E:

E с + E/C

Velocity of link F:

VF = VE + FC

The velocity image will be as below:

Vc C/B VBS VEC C VF/E4 VF

From the velocity diagram, measure the length of different vectors and angles to calculate velocities of various points:

V_{C}=0.619\,m/s\, \, \textrm{at}\, \, 22.7^{\circ}\, \, \textrm{north of west}

VC/B 0.851 m/s at 86.2° south of east

VE = 1.238 m/s at 7.6° north of west

VE/C = 0.660 0.660 m/s at 6.5° south of west

VF = 1.228 m/s horizontally towards west

VF/E = 0.164 m/s at 89.5° south of west

Calculate the normal or radial acceleration of B:

VB? П ав TAB

=\frac{1.256^{2}}{0.1}

= 15.77 m/s

As the normal acceleration is directed inwards to the link, the direction of this acceleration is 309 south of west.

Now calculate the normal or radial acceleration of C

{a_{C}}^{n}=\frac{{V_{C}}^{2}}{r_{CD}}

0.6192 03

= 1.28 m/s2

As the normal acceleration is directed inwards to the link, the direction of this acceleration is 67.39 south of west.

Calculate the normal or radial acceleration of E

{a_{E}}^{n}=\frac{{V_{E}}^{2}}{r_{DE}}

1.2382 06

= 2.55 m/s2

As the normal acceleration is directed inwards to the link, the direction of this acceleration is 車 82.4 south of west.

Calculate the normal or radial acceleration of C with respect to B

VCIB n ac/B TBC

0.8512 04

=1.81\,m/s^2

As the normal acceleration is directed inwards to the link, the direction of this acceleration is 3.8 south of west.

Calculate the normal or radial acceleration of E respect to C

Vec QE/C TCE

=\frac{0.66^{2}}{0.32}

= 1.361 m/s

As the normal acceleration is directed inwards to the link, the th direction of this  acceleration is 83.5^{\circ} south of east.

Now calculate the normal or radial acceleration of F with respect to E

2 VF/E 72 AFE TEF

=\frac{0.164^{2}}{0.65}

= 0.04 m/s

As the normal acceleration is directed inwards to the link, the direction of this  acceleration is 059 south of east.

Draw acceleration image to scale of the material handling mechanism:

JaE/C Race E/C VF/E af a af n Пар C/B n laC/B

From the acceleration daigram , measure the lengths of vector AF to calculate tangential acceleration of slider:

af = 34.43 m/s2 horizontally towards west

Now calculate the weight of piston

W = m x 9

= 4 x 9.8

= 39,2

Calculate the force acting on piston by virtue of its tangential acceleration

F = mx af

4 x 34.43

= 137.7 N

Now draw the free body daigram of piston

W F so F61ر F 65 F 61

Forces in horizontal direction:

ΣF, = 0

Fa + x F61 – F65 X cos 0.5º = 0

137.7 +0.15 × Fil Fas X Cos 0.5° = 0 +1

Forces in vertical direction:

\sum F_{y}=0

-w+ F_{61}-F_{65}\times \sin 0.5^{\circ}=0

-39.2+ F_{61}-F_{65}\times \sin 0.5^{\circ}=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \rightarrow 2

From the equation 2 into equation 1 :

137.7 + 0.15 x (F65 X sin 0.5° +39.2) - F65 X cos 0.5º = 0

F_{65}=143.6\,N

Now draw the free body diagarm of link 4 as shown below:

F45 FAT y 11

Consider moment equilibrium of link 4 about point D:

\sum M_{D}=0

F_{43}\times r_{1}-F_{45}\times r_{2}=0

F 43 X 0.5684 – 143.6 x 0.5954 = 0

F_{43}=318.6\,N

Now draw the free body diagarm of link 2

さつき X 21 21

Consider moment equilibrium of link 4 about point A:

\sum M_{A}=0

T_{21}-F_{23}\times r=0

T_{21}-318.6\times 0.0441=0

T_{21}=14.04\,N-m

Note: Hope you understand the solution, kindly upvote. Thank you

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