A beam of uniform cross-section and uniform mass-density of mass $20kg$ is supported at ends. A mass of $5kg$ is placed at a distance of $\frac{\text{L}}{5\text{m}} $ from one of its end. If beam is L m long, what are reactions of supports? In dealing with problems of equilibrium of a rigid body (like beam in this question). first of all draw a free body diagram of the system, indicating all of the forces acting on the system.
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Ia given case, our system looks like
Forces involved are:
In static equilibrium problem, calculation involves the following equations Sum of all vertical (or vertical component of) forces is zero, $\Sigma\text{F}_{\text{v}}=0$ Sum of all horizontal forces or horizontal components of forces is zero, $\Delta\text{F}_{\text{H}}=0$ Sum of all torques is zero. $\Sigma\tau=0$ You may take signs as All forces along + y-direction are positive. All forces along + x-direction are positive. All forces along - y-direction are negative. All forces along -x-direction are negative. All anti-clockwise torques are positive. All clockwise torques are negative. In given problem, as the beam is in equilibrium, $\Sigma\text{F}_{\text{v}}=0$
$\text{or }\text{F}_3+\text{F}_4-\text{F}_2-\text{F}_1=0$
$\text{or }\text{F}_36+\text{F}_4-\text{m}_2\text{g}=0\dots(1)$
$\Sigma\text{F}_{\text{H}}=0$ As there is no horizontal force involved here so, this does not give any equation. and $\Sigma\tau=0$ Let us choose rotation axis through right hand corner of beam.
Sum of all torques is $\text{F}_2\times\frac{4}{5}\text{L}+\text{F}_1\times\frac{\text{L}}{2}-\text{F}_3=0$ (no torque due to force F_4) From above equation, $\frac{4}{5}\text{F}_2+\text{F}_1-\text{F}_3=0$
$\text{or }\frac{4}{5}\text{m}_2\text{g}+\text{m}_1\text{g}-\text{F}_3=0\dots(2)$ When all the equations are formed, solve them like linear simultaneous equations to get desired results. In given problem, from Eq. (ii) we get $\text{F}_3=\frac{4}{5}\text{m}_2\text{g}+\text{m}_1\text{g}$
$=\frac{4}{5}\times5\times10+20\times10$
$=20\times12=240\text{N}$
$\text{m}_1=20\text{kg}$
$\text{and }\text{m}_2=5\text{kg}$
$\text{g}=10\text{m/s}^2$ Substituting value of $F_3, m_1$, and $m_2$_, in Eq. (i), we get $240+\text{F}_4-5\times10-20\times10=0$
$\text{or }\text{F}_4=50+200-240$
$\text{F}_4=10\text{N}$ So, support reactions are 240N on left support and 10N on right support.
Forces involved are:
- Weight of beam = $F_1 = m_1g$.
- Weight of mass placed over bean
- Reactions of supports F1 and Fa which we have to found.
In static equilibrium problem, calculation involves the following equations Sum of all vertical (or vertical component of) forces is zero, $\Sigma\text{F}_{\text{v}}=0$ Sum of all horizontal forces or horizontal components of forces is zero, $\Delta\text{F}_{\text{H}}=0$ Sum of all torques is zero. $\Sigma\tau=0$ You may take signs as All forces along + y-direction are positive. All forces along + x-direction are positive. All forces along - y-direction are negative. All forces along -x-direction are negative. All anti-clockwise torques are positive. All clockwise torques are negative. In given problem, as the beam is in equilibrium, $\Sigma\text{F}_{\text{v}}=0$
$\text{or }\text{F}_3+\text{F}_4-\text{F}_2-\text{F}_1=0$
$\text{or }\text{F}_36+\text{F}_4-\text{m}_2\text{g}=0\dots(1)$
$\Sigma\text{F}_{\text{H}}=0$ As there is no horizontal force involved here so, this does not give any equation. and $\Sigma\tau=0$ Let us choose rotation axis through right hand corner of beam.
Sum of all torques is $\text{F}_2\times\frac{4}{5}\text{L}+\text{F}_1\times\frac{\text{L}}{2}-\text{F}_3=0$ (no torque due to force F_4) From above equation, $\frac{4}{5}\text{F}_2+\text{F}_1-\text{F}_3=0$
$\text{or }\frac{4}{5}\text{m}_2\text{g}+\text{m}_1\text{g}-\text{F}_3=0\dots(2)$ When all the equations are formed, solve them like linear simultaneous equations to get desired results. In given problem, from Eq. (ii) we get $\text{F}_3=\frac{4}{5}\text{m}_2\text{g}+\text{m}_1\text{g}$
$=\frac{4}{5}\times5\times10+20\times10$
$=20\times12=240\text{N}$
$\text{m}_1=20\text{kg}$
$\text{and }\text{m}_2=5\text{kg}$
$\text{g}=10\text{m/s}^2$ Substituting value of $F_3, m_1$, and $m_2$_, in Eq. (i), we get $240+\text{F}_4-5\times10-20\times10=0$
$\text{or }\text{F}_4=50+200-240$
$\text{F}_4=10\text{N}$ So, support reactions are 240N on left support and 10N on right support.
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