Test 1 Memo PDF

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Past Semester test from 2017...

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Question 1 For the instant represented in the figure, car A has an acceleration in the direction of its motion, and car B has a speed of 15 m/s which is increasing. If the acceleration of B as observed from A is zero for this instant determine the acceleration of A and the rate at which the speed of B is changing. Use the angle θ = 45o and R = 100 m.

Vb m/s

Theta deg 15

Rm 45 0.7854

100

Calculate centripetal acceleration of B Ab = v^2 / R Ab = 15^2 / 100 Ab =

2.25 m/s^2

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Tan Theta = Tangent Acc B/ Sen Acc B Tan 45 = TangenT Acc B/ 2.25 Tangent Acc B =

2.25 m/s^2

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Calculate acc of A and B Cos theta = Cent acc B / Acc A and B Cos 45 = 2.25 / Acc A and B Acc A and B =

3.182 m/s^2

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Question 2

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In the design of a conveyor belt system, small metal blocks are discharged with a velocity of Va = 0.25 m/s on to a ramp by the upper conveyor belt shown in the figure. If the kinetic coefficient of friction between the blocks and the ramp is 0.3, calculate the angle Ө which the ramp must make with the horizontal so that the blocks will transfer without slipping to the lower conveyor belt moving at the speed of B = 0.3 m/s. The distance H = 1.8 m.

Va m/s 0.25

Mu

Vb m/s 0.3

H m

0.3

1.8

Calculate energy before Ep = mgh Ep = m g 1.8 Ep = 17.658 m J

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Ek = 0.5mv^2 Ek = 0.5 m 0.25^2 Ep = 0.03125 m J

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Calculate energy after Ek = 0.5mv^2 Ek = 0.5 * m * 0.3 * v^2 Ek = 0.045 m J

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Calculate Fu Fu = N mu Fu = mg Cos theta * 0.3 Fu = 2.943 m Cos theta N

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Wd = Fmu d Wd = 2.943 m Cos theta * (1.8 / Sin theta) Wd = 5.2974 m/ Tan theta  Energy before = energy after 17.658m + 0.03125m = 0.045m + 5.2974m/Tan theta 0.29330 theta = 3 rad 16.81 deg  [12]

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Question 3

The

two

springs,

each

of

stiffness

k = 1.6 kN/m, are equal length and undeformed when θ = 0o.

The three spheres

are on links of length R = 0.18 m and rotate about point O. If the mechanism is released from rest in the position θ = 20o, determine its angular velocity when θ = 0o. The mass of each sphere is 1.4 kg. treat the spheres as particles and neglect the masses of the light rods and springs.

k kN/m

theta Deg

R m

1.6

0

1600

0

0.18

Theta Deg 20 0.34906 6

mass kg 1.4

Calculate upstretched spring length R^2 = x^2 + y^2 L^2 = 0.18^2 + 0.18^2 L= 0.2546 m  Calculate stretched length calculate angle angle =( 180 - (90 + 20))/2 Angle = 35 deg 0.61086 5 Sin 35 / 0.18 = Sin 110 / L1 L1 = 0.2949 m Calculate change in spring length L1 = 0.2949 -0.2546 L1 = 0.0403 m  Calculate compressed length Calculate angle angle = ( 180 - (90 - 20))/2 Angle = 55 deg Page 3 of 4

0.95993 1 Sin 55 / 0.18 = Sin 70 / L2 L2 = 0.2065 m Calculate change in spring length L2 = 0.2546 - 0.2065 L2 = 0.0481 m  Calculate energy before Ep = mgh Ep = 1.4 g * (0.18 -0.18 Cos 20) Ep = 0.1491 J  Es1 = 0.5 k x^2 Es1 = 0.5 * 1600 * 0.0403^2 Es1 = 1.2993 J

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Es2 = 0.5 k x^2 Es2 = 0.5 * 1600 * 0.0481^2 Es2 = 1.8509 J

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Calculate energy after Ek = 0.5 m V^2 Ek = 3 ( 0.5 * 1.4 V^2) Ek = 2.1 V^2 J

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Energy before = Energy after Ep + Es1 +Es2 = Ek 0.1491 + 1.2993 + 1.8509 = 2.1 V^2 V= 1.2534 m/s Calculate angular velocity V = R omega 1.2534 = 0.18 Omega Omega 6.96333 Rad/ = 3 sec

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