Static Equilibrium of a 10m Rod with a Hanging Weight at 1.46m Distance

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SUMMARY

The discussion focuses on the static equilibrium of a 10-meter rod weighing 50 Newtons, supported by a cable at a 53-degree angle and resting against a wall with a friction coefficient of 2. The calculations provided determine that a block weighing 20 Newtons can be hung at a minimum distance of 1.46 meters from the left end of the rod without causing it to slip. The equations used include force balance in both the x and y directions, as well as torque equilibrium, confirming the accuracy of the solution.

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nameVoid
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one end of a uniform 10 meter long rod weighing 50 Newtons is supported by a cable at an angel of 53 deg with the rod the other end rests against the wall where is is held by a friction us=2 . determine the minimum distance x from the left end of the rod at which a block weighing 20 Newtons can be hung without causing the rod to slip.

IS THIS CORRECT?

FX=n-Tcos53=0
FY=Tsin53+FS-70=0
FT=10Tsin53-250-20x=0
20x=10Tsin53-250
2x=Tsin53-25
FS=2Tcos53
Tsin53+2Tcos53=70
T=70/ sin53+2cos53
2x=70sin53/ sin53+2cos53 -25
x=35sin53/ sin53+2cos53 -25/2=1.46m
 
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nameVoid said:
one end of a uniform 10 meter long rod weighing 50 Newtons is supported by a cable at an angel of 53 deg with the rod the other end rests against the wall where is is held by a friction us=2 . determine the minimum distance x from the left end of the rod at which a block weighing 20 Newtons can be hung without causing the rod to slip.

IS THIS CORRECT?

FX=n-Tcos53=0
FY=Tsin53+FS-70=0
FT=10Tsin53-250-20x=0
20x=10Tsin53-250
2x=Tsin53-25
FS=2Tcos53
Tsin53+2Tcos53=70
T=70/ sin53+2cos53
2x=70sin53/ sin53+2cos53 -25
x=35sin53/ sin53+2cos53 -25/2=1.46m
It looks good to me. :approve:
 

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