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TUTORIAL 2

1. A force of F= 75 N is applied to the cord. Determine how high the 15

kg block A rises in 2 s starting from rest. The weight of pulleys is 2 kgand neglects the weight of the cord.

SOLUTION:

- Kinematic equations

12 LSS CB 0202 CB

CB vvdt

dS

dt

dS

0202 CBCB aa

dt

dv

dt

dv(1)

2)( LSSS CAA 0202 CA

CA vv

dt

dS

dt

dS

0202 CACA aa

dt

dv

dt

dv(2)

- Free body diagram

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- Kinetics Equation

Blok A

AAPA amWWTF 12

AaTF 15)81.9(2))81.9(152 1AaT 5.777.1661 (3)

Pulley C CPP amWTTF 122

CA aaT 2)81.9(2)5.777.166(2 2 (4)

Pulley B 752 TF

Substitute the value ofT2 into equation (4)

CA

CA

aa

aa

25.739.36

2)81.9(25.777.166()75(2

Combine with equation (2) 02 CA aa

smaaa AAA /16.3)2(25.739.36 (direction upward)

smaC /32.6

Position of block A after 2 s from rest 2

02

1attvSS oA

mSA 4.142)6.3(2

100

2

Position of pulley C after 2 s from rest mSC 8.282)32.6(2

100

2

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2. If the 5 kg blockA slides down the plane with a constant velocity when = 30o, determine the

acceleration of block A when = 45o.

Solution:

- Free body diagram

- Equation of motion:

030cosWNFy (no movement in y-direction)030cos)81.9(5 N NN 48.42

030sinWNFx (no movement in x-direction at =30)030sin)81.9(5)48.42( 58.0

- When = 45o

xx maWNF 30sinxa545sin)81.9(5)48.42(58.0

2/01.2 smax

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3. If a horizontal force of P = 50 N is applied to block A, determine the acceleration of block B.

Neglect the friction.

Solution

- Movement block A is related to movement block B. 15tanAB SS 15tanAB vv

15tanAB aa

AB aa 27.0 (1)

- Free body diagram

- Equation of motion

015cos21 AyA WNNF (Block A has no movement in y-direction)24.3997.0

21 NN (2)

AAxA amNPF 15sin2AaN 426.050 2 (3)

BBByB amWNF 15cos2BaN 5.756.7397.0 2 (4)

015sin 32 NNFxB (Block B has no movement in x-direction)3226.0 NN (5)

Substitute equation (1) to (3) and combine with equation (4)

97.0

75

)27.0(26.0

4

97.0

56.73

26.0

50Ba

2/87.0 smaB