A rod 7.0 m is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, what distance from the pivot can a third force of 300 N acting upward be placed to produce rotational equilibrium? Neglect the weight of the rod.
1.0, 2.0,3.0, or 4.0 ?
2006-12-05 14:55:26 · 2 answers · asked by Kitana 2 in Science & Mathematics ➔ Physics
The only logical answer of the 4 would be 1.0. I only say this because if you look at having the same weight (50g) then at 2 vs 2, it would be even. With 2 on the left, 3 or 4 on the right, the right would weigh down. Therefore since the left is lighter (50g) vs. the right(300g), then the only choice of possibly balancing it would be at the 1.0 distance.
2006-12-05 15:05:44 · answer #1 · answered by arj7002 2 · 0⤊ 0⤋
Your question seems incomplete. We are to neglect the weight of the rod which therefore generates no force. We have one force of 50 N acting on the left end. You ask where a THIRD force should be applied. I didn't find the SECOND force. Without knowing this the problem cannot be solved.
Bramble
2006-12-05 16:10:17 · answer #2 · answered by Bramble 7 · 0⤊ 0⤋