PHYSICS HELpp?

Question

The following is a link to the diagram containing the problem:
http://i7.photobucket.com/albums/y281/cowmoocher/IMG.jpg

Thanks in advance :}

Answer 1

Assume the pulley is mass less
and the cord does not slip
T1 is the tension in the cord. And a is
the acceleration upward of A and downward of C
Look at FBDs of each mass
start with the combined A and B

T1-3*g=3*a

combined C and D
7*g-T1=7*a

solve for T1 and a

T1/3-g=a

7*g-T1=7*T1/3-7*g
14*g=10*T1/3
using g=10

T1=42 N
This is part 2 of the question


1) The acceleration of mass D:
D will accelerate downward at a
the magnitude of a is 4 m/s^2
for D a=-4 m/s^2

3) A FBD of D shows
4*10-T2=4*4
T2=40-16
T2=24 N

4) A FBD of B shows
F=2*10+2*4
F=28 N

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