(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the mi
ddle point. On the right hand side, a weight of 200 g is suspended. When the stick is balanced, where should the 200 g weight be suspended (how many cm from the middle point)?
1 answer:
Answer:
20cm
Explanation:
Hello!
remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,
To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation
(100g)(40cm)=x(200g)
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Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.
So,
F =
Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.
Let's suppose, q1 = q
So,
if q1 = q
then
q2 = Q-q
Product of Charges = q1 x q2
Now, it is:
Product of Charges = q x (Q-q)
So,
Product of Charges = qQ -
And the expression qQ - is clearly a quadratic expression. And clearly its roots are 0 and Q.
So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.
So, the midpoint is:
q =
q = Q/2 and it is the highest value of each charge in order to get the greatest force.
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