Next: , Previous: , Up: Examples dynamics of particles   [Contents][Index]

11.11 Example 11

An A block of 23 kg, is above a B block of 17 kg, as show the image. If the friction coefficient in all surfaces is 0.18, and F force is of 235 Newtons, which is the acceleration of the block A? Which is the stress in the string? Solution with FisicaLab

Erase the content of the chalkboard and select the SI system. And add the necessary elements to make the free body diagram of each block. As show the image below: We assume that the system leaves the rest, and assume a time of 0.5 seconds. Then to the element Stationary reference system write:

g

9.81

t

0.5

To the block that correspond with the A block, write:

Name

A

m

23

ang

25

a

aA

vi

0

vf

vfA

d

dA

Relative to

sf

To its normal:

f

nA

ang

65

To the element Friction between blocks write:

N

nA

u

0.18

ang

25

To the force that correspond with force applied to the string, write:

f

t

ang

25

Now, to the B block, write:

Name

B

m

17

ang

25

a

aB

vi

0

vf

vfB

d

dB

Relative to

sf

To its normal:

f

nB

ang

65

To the friction with the plane:

N

nB

u

0.18

To the friction with A block, the element Friction between blocks, write:

N

nA

u

0.18

ang

25

To the applied force of 235 Newtons, write:

f

235

ang

25

To the force that correspond with force applied to the string, write:

f

t

ang

25

To the force applied to the A block (the normal reaction):

f

nA

ang

65

And to the relation between the accelerations:

a1

aA

a2

aB

z

-1

Now click in the icon Solve to get the answer:

nB = 355.635 N ;  t = 173.848 N ;  nA = 204.490 N ;
aA = 1.812 m/s2 ;  aB = -1.812 m/s2 ;  vfB = -0.906 m/s ;
dB = -0.227 m ;  vfA = 0.906 m/s ;  dA = 0.227 m ;
Status = success.

WARNING: Check that the sense of friction forces is
opposite of the movement sense (when correspond).

Next: , Previous: , Up: Examples dynamics of particles   [Contents][Index]