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

AI Thread Summary
The discussion focuses on determining the minimum distance from the left end of a 10-meter rod, supported by a cable and resting against a wall, at which a 20 Newton block can be hung without causing the rod to slip. The calculations involve analyzing forces in both the x and y directions, using tension and friction equations. The final result suggests that the block can be hung at a distance of 1.46 meters from the left end of the rod. Participants confirm the calculations appear correct. The thread emphasizes the importance of static equilibrium in solving the problem.
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:
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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