The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent ho
1 answer:
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Answer:
= 14.88 N
Explanation:
Let's begin by listing out the given variables:
M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,
g = 9.8 m/s²
At equilibrium, the sum of all external torque acting on an object equals zero
τ(net) = 0
Taking moment about we have:
(M + m) g * 0.5L - (L - d) = 0
⇒ = [(M + m) g * 0.5L] ÷ (L - d)
= [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)
= 59.535 ÷ 2.4
= 24.80625 N ≈ 24.81 N
Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N
Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N
Using sum of equilibrium in the vertical direction, we have:
+
= W + w ------- Eqn 1
Substituting T2, W & w into the Eqn 1
+ 24.81 = 26.46 + 13.23
= <u>14.88</u> N
Answer:
Explanation:
Answer:
0.632 m
Explanation:
let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.
According to the diagram,
d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)
Let T be the tension in the string.
Resolve the components of T.
T Sinθ = k q1 q2 / d^2
T Sinθ = k q1 q2 / (2LSinθ)² .....(2)
T Cosθ = mg .....(3)
Dividing equation (2) by equation (3), we get
tanθ = k q1 q2 / (4 L² Sin²θ x mg)
tan θ Sin²θ = k q1 q2 / (4 L² m g)
For small value of θ, tan θ = Sin θ
So,
Sin³θ = k q1 q2 / (4 L² m g)
Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)
Sin³θ = 0.2523
Sinθ = 0.632
θ = 39.2 degree
So, the separation between the two charges, d = 2 x L x Sin θ
d = 2 x 0.5 x 0.632 = 0.632 m
