dynamics tutorials 7 and 8 on kinetics 3. Determine the initial acceleration of the 10-kg smooth collar. The spring has an unstretched length of 1 m. PDF

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MMB241 Dynamics ofParticlesZeleke Migbar Assefa (PhD) E-mail: [email protected] migbarassefa@gmail Office: 143 (Building 248)####### Tutorial 7####### Kinetics (EOM) The 10-kg block is subjected to the forces shown. In each case, determine its velocity when t = 2 s if v = 0 when t = 0. a) b)98N= 10aFBD ...

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MMB241 Dynamics of Particles Tutorial 7 Kinetics (EOM) Zeleke Migbar Assefa (PhD) E-mail: [email protected] [email protected] Office: 143 (Building 248)

1. The 10-kg block is subjected to the forces shown. In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

a)

b)

a) 98.1N

= FBD

F

x

N

10a KD

 ma x ; 500 cos  300  ma x

4 500   300  10 a x  a x  a  10 m / s 2 5 v  u  at  v  0  10(2)  20 m / s

b) 98.1N

= FBD

KD

N

F

x

10a

 ma x ; 20t  10a a  2t

dv  adt;

v

2

0

0

 dv   2t dt  v  4m / s

2. The 10-kg block is subjected to the forces shown. In each case, determine its velocity at s = 8 m if v = 3 m/s at s = 0. Motion occurs to the right.

a)

b)

(a)

Velocity?? at s = 8 m if v = 3 m/s at s = 0

=

98.1N

FBD

KD

N

F

x

10a

 ma x ; 40  30  10 a a  1m / s 2

vdv  ads

v 2  vo2  2ac (s  so ) v 2  32  2(1)(8  0)  v  5m / s

98.1N

(b)

Velocity?? at s = 8 m if v = 3 m/s at s = 0

= FBD

2.5s  10a a  0.25s vdv  ads

KD

N

 Fx  max 

10a

v

8

3

0

 vdv   0.25s ds v 2  32  2(0.125  82 )  v  5m / s

3. Determine the initial acceleration of the 10-kg smooth collar. The spring has an unstretched length of 1 m.

Fs  kx  (10)(5  1)  40 N

F

x

 ma x ;

4 (40)  10a  a  3.2m / s 2 5

4. Write the equations of motion in the x and y directions for the 10-kg block.

N

10(9.81)cos30 10*9.81 10(9.81)sin30

FBD

98.1sin 30

98.1cos 30

 Fx  max ;

98.1sin 30  0.2 N  10a

 Fy  may ;

N  98.1cos 30  0

Repeated

5. The motor winds in the cable with a constant acceleration, such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. Determine the tension developed in the cable. The coefficient of kinetic friction between the crate and the plane is μk = 0.3.

y

F

x

x

 max ;

T  Fs  mg sin   ma N

T

Ff

30

mg

1 s  s o  vo t  ac t 2 2 1 6  0  0  a (32 ) 2 a  1.333m / s 2

F

y

 may ;

N  mg cos   0 N A  20(9.81) cos 30  0

F

x

 max ;

N A  169.91N

T  Fs  mg sin   ma  T  0.3  169.9  20(9.81) sin 30   20(1.333)  176 N

6. If motor M exerts a force of F = (10t2 + 100) N on the cable, where t is in seconds, determine the velocity of the 25-kg crate when t = 4 s. The coefficients of static and kinetic friction between the crate and the plane are μs = 0.3 and μk = 0.25, respectively. The crate is initially at rest.

  Ff

max

  s N A  0.3(245.25)  73.575N

F  100 N  ( F f ) max

when t= 0, the crate will start to move immediately after F is applied..

F

N

y

T

 ma y ;

N A  25(9.81)  0 N A  245.25 N

F

x

 max ;

10t 2 100  k N  ma

Ff

mg

10t 2  100  0.25(245.25)  25 a a  (0.4t 2  1.5475) v

4s

0

0

dv  adt   dv   (0.4t 2  1.5475)dt v  14.7 m / s 

7. A spring of stiffness k = 500 N/m is mounted against the 10-kg block. If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and the spring is uncompressed. The contact surface is smooth.



Fx  max ;

F cos   Fs  ma

4 500   500s  10a 5 400  500 s  10 a

N

a  40  50 s ads  vdv Fs

mg 40s  25s

2 0.5 0

v2  2

v 0

0.5

v

0

0

 (40  50s )ds   vdv

 v  2(40  0.5)  25(0.5) 2  5.24 m / s...