 #1
jkfjbw
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 Homework Statement:
 A horizontal bar of length ##L## and mass ##m_B## is on two fulcrums, one ##L/3## from the left side of the bar, and the other ##L/3## from the right side of the bar. There is also a mass of mass ##m_D## on the left end of the bar. Find the forces that the fulcrums exert on the bar.
 Relevant Equations:

$$\sum_i \vec{F}_i = \vec{0}$$
$$\sum_i \tau_{z,i} = 0$$
A diagram of the physical situation is below:
Choosing the positive ##y## direction to be upwards and the positive direction of rotation to be counterclockwise, Newton's linear second law gives:
$$m_D g + F_L + F_R = 0$$
where ##F_L## is the magnitude of the force exerted on the bar by the left fulcrum, and ##F_R## is the magnitude of the force exerted on the bar by the right fulcrum. Taking the point of rotation to be around the left fulcrum, Newton's rotational second law gives:
$$(m_D g)\Big(\frac{L}{3}\Big) + F_R\Big(\frac{L}{3}\Big)  (m_B g)\Big(\frac{L}{6}\Big) = 0\\
m_D g + F_R  \frac{1}{2} m_B g = 0\\
F_R = g\Big(\frac{1}{2}m_B  m_D\Big)$$
which is the correct answer for ##F_R##. The problem comes when I try to find ##F_L## by using Newton's linear second law to eliminate ##F_R## and solve for ##F_L##. Doing so gives:
$$m_D g + m_D g  F_L  \frac{1}{2}m_B g = 0\\
F_L = 2m_D g  \frac{1}{2}m_B g\\
F_L = g\Big(2m_B  \frac{1}{2}m_B\Big)$$
which is the wrong answer for ##F_L##.
Likewise, I can also take the point of rotation to be around the right fulcrum:
$$(m_D g)\Big(\frac{2}{3}L\Big)  F_L\Big(\frac{L}{3}\Big) + (m_B g)\Big(\frac{L}{6}\Big) = 0\\
2m_D g  F_L + \frac{1}{2}m_B g = 0\\
F_L = g\Big(2m_D + \frac{1}{2}m_B\Big)$$
which is the correct answer for ##F_L##. However, if I use Newton's linear second law to eliminate ##F_L## and solve for ##F_R##, I get the wrong answer:
$$2m_D g  m_D g + F_R + \frac{1}{2}m_B g = 0\\
m_D g + F_R + \frac{1}{2}m_B g = 0\\
F_R = g\Big(m_D + \frac{1}{2}m_B\Big)$$
Why do I get the wrong answer for ##F_L## and ##F_R## when I substitute Newton's linear second law into the knowncorrect answer for ##F_R## and ##F_L##?
Choosing the positive ##y## direction to be upwards and the positive direction of rotation to be counterclockwise, Newton's linear second law gives:
$$m_D g + F_L + F_R = 0$$
where ##F_L## is the magnitude of the force exerted on the bar by the left fulcrum, and ##F_R## is the magnitude of the force exerted on the bar by the right fulcrum. Taking the point of rotation to be around the left fulcrum, Newton's rotational second law gives:
$$(m_D g)\Big(\frac{L}{3}\Big) + F_R\Big(\frac{L}{3}\Big)  (m_B g)\Big(\frac{L}{6}\Big) = 0\\
m_D g + F_R  \frac{1}{2} m_B g = 0\\
F_R = g\Big(\frac{1}{2}m_B  m_D\Big)$$
which is the correct answer for ##F_R##. The problem comes when I try to find ##F_L## by using Newton's linear second law to eliminate ##F_R## and solve for ##F_L##. Doing so gives:
$$m_D g + m_D g  F_L  \frac{1}{2}m_B g = 0\\
F_L = 2m_D g  \frac{1}{2}m_B g\\
F_L = g\Big(2m_B  \frac{1}{2}m_B\Big)$$
which is the wrong answer for ##F_L##.
Likewise, I can also take the point of rotation to be around the right fulcrum:
$$(m_D g)\Big(\frac{2}{3}L\Big)  F_L\Big(\frac{L}{3}\Big) + (m_B g)\Big(\frac{L}{6}\Big) = 0\\
2m_D g  F_L + \frac{1}{2}m_B g = 0\\
F_L = g\Big(2m_D + \frac{1}{2}m_B\Big)$$
which is the correct answer for ##F_L##. However, if I use Newton's linear second law to eliminate ##F_L## and solve for ##F_R##, I get the wrong answer:
$$2m_D g  m_D g + F_R + \frac{1}{2}m_B g = 0\\
m_D g + F_R + \frac{1}{2}m_B g = 0\\
F_R = g\Big(m_D + \frac{1}{2}m_B\Big)$$
Why do I get the wrong answer for ##F_L## and ##F_R## when I substitute Newton's linear second law into the knowncorrect answer for ##F_R## and ##F_L##?
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