Question

The 450-kg uniform I-beam supports the load of 220 kg as shown.
Determine the reactions at the supports.

R = ?
R = ?
5.6 m
2.4 m
220 kg​

Answer 1

Answer:

2850 N and 3720 N

Explanation:

Answer 2

Concept Introduction:-

In gymnastics, a beam balance is a narrow wooden beam hanging in a horizontal position around it's four feet off the ground and is utilised for balancing tricks.

Given Information:-

We have been given that the $450$-kg uniform I-beam supports the load of $220$ kg.

To Find:-

We have to find that the Force of Reaction $=R A=$ ? and Force of Reaction $=R B=$ ?

Solution:-

According to the problem

Mass of l-beam $=m b=450 \mathrm{~kg}$

Weight of l-beam $=W b=F b=m g=450 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2}=4410 \mathrm{~N}$

Mass of the load $=m l=220 \mathrm{~kg}$

Weight of load $=W l=F l=m g=220 \mathrm{~kg} \times 9.81 \mathrm{~m} \mathrm{~s}^{-2}=2156 \mathrm{~N}$

First we will apply the first condition of equilibrium, which is mathematically represented as

$\Sigma F=0$

$\Sigma F( upwards )=\Sigma F( downwards )$

$\begin{aligned}&R A+R B=F b+F l \\&R A+R B=4410 N+2156 N\end{aligned}$

$R A+R B=6566 \mathrm{~N}$

Now considering moments about point A. then

Moment arm for $\mathrm{FL}=r \mathrm{~L}=5.6 \mathrm{~cm}$

Length of l-beam $=2.4 \mathrm{~cm}+5.6 \mathrm{~cm}=8 \mathrm{~cm}$

Moment arm for $R B=r B=8 \mathrm{~cm}$

Moment arm for $F b=r c g=L / 2=4 / 2=2 \mathrm{~cm}$

By applying the second condition of equilibrium; which is mathematically written as $\Sigma T=0$

or

$\Sigma$ T Anti-clock wise $=\Sigma$ T Clock wise

$r b \times R B=r b \times F b+r L \times R L$

$8 \mathrm{~cm} \times R B=4 \mathrm{~cm} \times 4410 \mathrm{~N}+5.6 \mathrm{~cm} \times 2156 \mathrm{~N} \mathrm{c}$

$8 \mathrm{~cm} \times R B=17,640 \mathrm{~N} \mathrm{~cm} \times 12,073.6 \mathrm{~N}$

$8 \mathrm{~cm} \times \mathrm{RB}=29,713.6 \mathrm{~N}$

$\begin{aligned}&R B=29,713.6 \mathrm{~N} \mathrm{~cm} / 8 \mathrm{~cm} \\&R B=3,714.2 \mathrm{~N}\end{aligned}$

or

$R B=3,714 \mathrm{~N}$

Putting value of $R B=3,714 \mathrm{~N}$ in equation $(1)$, we get

$\begin{aligned}&R A+3,714 N=6566 \mathrm{~N} \\&R A=6566 \mathrm{~N}-3,714 \mathrm{~N} \\&R A=2,852 \mathrm{~N} \cong 2850 \mathrm{~N}\end{aligned}$

Thus,

Force of Reaction $=\mathrm{RA}=2,850 \mathrm{~N}$

Force of Reaction $=\mathrm{RB}=3,714 \mathrm{~N}$

Final Answer:-

The value of $RA$ and $RB$ is $2850$ N and $3714$ N respectively.

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