Physics, 09.10.2019 11:20 iibabycarrotsii

Derive an algebraic expression for the tension in the segment of the cord between the post and block a (assumed massless). express your answer in terms of the variables ma, mb, ra, rb, f, and appropriate constants.

Derive an algebraic expression for the tension in the segment of the cord between the post and block

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Answer from: P4thing

T = 4· $\pi ^{2}$ ·f (mA · rA + mB · rB)

Explanation: According to Newton's First Law of motion, an object stays in the same state of motion unless a resultant force acts on it. If the resultant force is zero, the object stays stationery or in movement with the same velocity. In the system described, the two blocks are moving in the same frequency (f). For the block A, the forces acting on it are Centripetal Force on block A (FcA), Tension of the cord and Centripetal Force of block B (FcB). This last one counts because the two blocks are connected to each other with the same cord. So, the Resultant Force will be

FcA+ FcB + T = 0

Now, Centripetal Forces always points to the center of the circular movement.

Assuming that pointing towards the center is positive, the expression will be: FcA+ FcB - T = 0, in which tension (T) acts opposite to the others.

To calculate Fc, the expression is

Fc = m · ac

The term ac is the centripetal acceleration and is expressed as

ac = $\frac{v^{2} }{R}$ and can also be written as ac = ω²·R.

As ω is the angular velocity and can be written as ω = 2· $\pi$ ·f, we have

ac = (2· $\pi$ ·f)²·R

So, FcA+ FcB - T = 0

Calculating and substituting:

- T = - FcA - FcB

T = FcA + FcB

T = mA·(2· $\pi$ ·f)²·rA· + mB·(2· $\pi$ ·f)²·rB

T = 4· $\pi ^{2}$ ·f (mA · rA + mB · rB)

Answer from: Quest

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Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge