Question

A laboratory on the concept of static equilibrium uses an apparatus called a force table. The force table consists of a round, level platform marked in degree increments about its circumference. In the center of the table sits a massless ring, to which three strings are tied. Each string is strung over a separate pulley clamped to the edge of the platform, and then tied to a freely-hanging mass, so that each string is under tension and the ring is suspended parallel to the table surface. A mass of mı = 0.165 kg is located at 8, = 28.5°, and a second mass of my = 0.219 kg is located at 02 = 295 Calculate the mass mz and the angular position (in degrees) that will balance the system and hold the ring stationary over the center of the platform. m3 = 0g =

Answer 1

Given:

mı = 0.165kg at 01 = 28.5

m1 = miri+m1yj = micosiji + misin01)

$m_{1}=0.165kg*cos28.5i+0.165kg*sin28.5j$

m = 0.1451 +0.07871

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Given:

m2 = 0.219kg at 0 = 295

$m_{2}=m_{2x}i+m_{2y}j=m_{2}cos\theta _{2}i+m_{2}sin\theta _{2}j$

$m_{2}=0.219kg*cos295i+0.219kg*sin295j$

${\color{Red} m_{2}=0.0926i-0.1985j}$

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Given that the system is in equilibrium

So $m_{1}+m_{2}+m_{3}=0$

$(0.145i+0.0787j)+(0.0926i-0.1985j)+m_{3}=0$

$0.145i+0.0787j+0.0926i-0.1985j+m_{3}=0$

$(0.145+0.0926)i+(0.0787-0.1985)j+m_{3}=0$

$0.2376i-0.1198j+m_{3}=0$

${\color{Red} m_{3}=-0.2376i+0.1198j}$

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Magnitude

$m_{3}=\sqrt{(-0.2376)^{2}+(0.1198)^{2}}$

ANSWER: ${\color{Red} m_{3}=0.2661kg}$

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Direction

$\theta _{3}=tan^{-1}(\frac{m_{3y}}{m_{3x}})$

$\theta _{3}=tan^{-1}(\frac{0.1198}{-0.2376})$

$\theta _{3}=-26.76^{\circ}$ (negative x and positive y. So 2nd quadrant)

$\theta _{3}=180^{\circ}-26.76^{\circ}$

ANSWER: ${\color{Red} \theta _{3}=153.24^{\circ}}$

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