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1 year ago

What is the net torque on the bar shown in (Figure 1) about the axis indicated by the dot? Suppose that F=6.0N

Physics

1 answer:

AysviL [449]1 year ago

6 0

Answer:

net torque on the system is 2.5 Nm

Explanation:

As we know that net torque on the system is given as

$\tau = r_1F_1 - r_2F_2$

now we have

$r_1 = 75 cm$

$F_1 = 6 N$

$r_2 = 25 cm$

$F_2 = 8 N$

now we have

$\tau = 0.75 (6) - 0.25(8)$

$\tau = 2.5 Nm$

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Scorpion4ik [409]

Answer:

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Explanation:

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Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

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Answer:

Two possible points

x= 0.67 cm to the right of q1

x= 2 cm to the left of q1

Explanation:

Electrostatic Forces

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$