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Hanging masses and tensions

  • Thread starter c-murda
  • Start date
  • #1
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[SOLVED] Hanging masses and tensions....

Homework Statement


attached is drawing of diagram...

two masses m1 and m2 are attached as shown in the figure. If the system is in equilibrium, find the angle labled theta.
m1 = 35.0 kg
m2 = 65.0 kg




Homework Equations





The Attempt at a Solution


well...to start i drew three FBD one for the mass, one for the intersection, and one for the second mass...
for the first FBD:
Tm = m * g

for the second FBD:
Fx = T2 cos 53 - T1 cos 37 = 0
Fy = T2 sin 53 + T1 sin 37 - m1*g = 0

for the third FBD:
Fx = T3 cos (theta) - T2 cos 53 = 0
Fy = T3 sin (theta) - T2 sin 53 - m2*g = 0

so...

First:
Tm = m*g; which equals (35.0kg) ( 9.80 m/s^2) = 343N

Second:
.602 T2 - .799 T1 = 0 ; so... T1 = (.602/.799)(T2) so...

.602 T1 +.799 T2 = 343
.602[((.602)/(.799))T2 + .799 T2 = 343
.453T2 + .799T2 = 343
1.25T2 = 343
T2 = 274.4N

Third:
T3=T2(cos 53/cos (theta))

[T2(cos 53/cos (theta))] sin(theta) - T2 sin 53 = m2*g
T2 tan(theta) cos 53 = m2*g + T2 sin 53
tan(theta) = m2*g + T2 sin 53/T2 cos 53
theta = tan^-1(m2*g + T2 sin 53/T2 cos 53)
theta = 79.1?


uhmm any help? thats what i got correct me if i am wrong
 

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Answers and Replies

  • #2
67
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anyone?
 
  • #3
Doc Al
Mentor
44,945
1,206
Looks good to me. Well done!
 
  • #4
67
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thank you much...i have more to come so sit tight!
 

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