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A rigid rod is pivoted at point O and caries a body of weight w1 at the left end A.Find...
A rigid rod is pivoted at point O and caries a body of weight w1 at the left end A.
Find the weight w2 of another body which must be attached at right end B if the rod is to be in equilibrium and find the force exerted on the rod by the pivot at O. The weight of the rod is negligible.
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Since the rod is in equilibrium, the net force and the net torque on it must be zero.
The forces T1 and T2 are acting in opposite dirrection to the force P.
P is the force of the pivot on the rod.
We'll write the equilibrium equations:
T1 + T2 - P = 0
T1 + T2 = P (1)
The point O is located at the length l1 from the left end A and at the length l2 from the right end B.
T2*l2 - T1*l1 = 0
T2*l2 = T1*l1 (2)
We'll divide by l2:
T2 = T1*l1/l2
We'll substitute (2) in (1) and we'll get:
T1 + T1*l1/l2 = P
We'll factorize by T1 and we'll get:
T1*(1 + l1/l2) = P
We know that T1 = W1
W1*(1 + l1/l2) = P
T2 = T1*l1/l2 and since T2 = W2, we'll get:
W2 = W1*l1/l2
Posted by giorgiana1976 on October 25, 2010 at 4:14 PM (Answer #1)
The arrangement described in the question forms a lever system in which the the fulcrum or the pivot is in the center with the load (w1) and the effort (w2) are on opposite side of the fulcrum.
A system like this is in equilibrium when:
Load x (Distance of load from fulcrum) = Effort x (Distance of effort from fulcrum)
Therefore, for the given question:
w1 x (Distance of w1 from pivot) = w2 x (Distance of w2 from pivot)
w2 = w1 x [(Distance of w1 from pivot)/(Distance of w2 from pivot)]
Further, both the load and effort will act in the same direction. Therefor. the total force on the pivot will be the sum of force exerted by the sum of two weights.
Force acting on the pivot = (w1 + w2) x (acceleration due to gravity)
Posted by krishna-agrawala on October 25, 2010 at 5:21 PM (Answer #2)
Here the point at which the rod is pivoted is not given. Let us take the length of the rod to be L and the pivot O is at a point that is x away from the left end. Therefore for equilibrium we need to hang a weight w2 at the right end and it will be L-x away from the pivot.
Now for equilibrium, the net torque at the pivot has to be equal to zero.
Therefore w1*x = w2* ( L-x)
=> w2 = w1*x / (L-x)
So the mass w2 has to be w1*x / (L-x)
The total force acting on the rod at the pivot is w1 + w2.
Posted by william1941 on October 25, 2010 at 4:15 PM (Answer #3)
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