A rod is 6 m long pivoted at a point 1.5 m from the left end. Two dow

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A 6 m rod is pivoted 1.5 m from the left end, with downward forces of 50 N and 200 N applied at each end. To maintain rotational equilibrium, a third upward force of 300 N must be applied at a specific distance from the pivot. The calculated distance for this force is 2.75 m from the pivot point. Participants confirm that this answer is correct, indicating no errors in the solution. The discussion emphasizes the importance of verifying calculations in physics problems.
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Homework Statement



A rod is 6 m long pivoted at a point 1.5 m from the left end. Two downward forces of
magnitude 50 N and 200 N are exerted on the left and right end of the rod
respectively. At what distance from the pivot point must a third upward force of
magnitude 300 N be exerted on the rod to keep it in rotational equilibrium? Neglect
the weight of the rod.



Homework Equations





The Attempt at a Solution



so apparently the answer for this question is "non of the above ".. I got an answer of 2.75, can someone check it for me please?
 
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meaning i got it right?
 
cmkc109 said:
meaning i got it right?

Yes. No straight A's for wrong answers here! :smile:
 
rude man said:
Yes. No straight A's for wrong answers here! :smile:

then can u show me how to get the right ans?
 
cmkc109 said:
then can u show me how to get the right ans?

rude man said you got the right answer. (And I agree.)
 
lol ok thanks!
 
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