Question

A lamina is in the shape of the isosceles trapezium The two parallel sides are 5 metre apart and length are 3 m and 2m. The distance of the centre of mass of gravity of trapezium from the longer parallel side

Answer 1

$\frac{7}{3}$

Explanation:

Y is the coordinate of centre of mass ,

$y_{com}= \frac{a_1y_1+a_2y_2+a_3y_3}{a_1+a_2+a_3}$

Area of triangle ABC, $a_1$=Area of triangle DEF, $a_2$

$= \frac {5}{4}m^2$

Area of rectangle ACDE , $a_3$  $=2\times5=10m^2$

y cordinate of centre of mass of a triangle is $\frac{h}{3}$

h = 5m

$y_1 = y_2 =\frac{5}{3}$

$y_3=\frac{5}{2}$

$y_{com}=\frac{7}{3}$

Answer 2

The distance of center of mass  is $\dfrac{7}{3}$

Explanation:

Given that,

First parallel side = 5 m

Second parallel side = 3 m

Length = 2 m

We need to calculate the area of triangle  

Using formula of area of triangle

Triangle ABF,

$A_{1}=\dfrac{}{}\times\dfrac{h}{2}$

$A_{1}=\dfrac{1}{2}\times\dfrac{1}{2}\times 5$

$A_{1}=\dfrac{5}{4}$

Triangle CDE,

$A_{2}=\dfrac{1}{2}\times\dfrac{1}{2}\times 5$

$A_{2}=\dfrac{5}{4}$

We need to calculate the area of rectangle BCEF,

Using formula of area of rectangle

$A_{3}=side\times\ height$

$A_{3}=2\times5$

$A_{3}=10\ m^2$

We need to calculate the y coordinate of center of mass of a triangle

Using formula of y coordinate of center of mass

$y=\dfrac{h}{3}$

Put the value into the formula  

$y_{1}=y_{2}=\dfrac{5}{3}$

We need to calculate the y coordinate of center of mass of a rectangle

Using formula of y coordinate of center of mass

$y=\dfrac{h}{2}$

Put the value into the formula

$y_{3}=\dfrac{5}{2}$

We need to calculate the distance of center of mass

Using formula of center of mass

$y=\dfrac{A_{1}y_{1}+A_{2}y_{2}+A_{3}y_{3}}{A_{1}+A_{2}+A_{3}}$

Put the value into the formula

$y_{cm}=\dfrac{\dfrac{5}{4}\times\dfrac{5}{3}+\dfrac{5}{4}\times\dfrac{5}{3}+10\times\dfrac{5}{2}}{\dfrac{5}{4}+\dfrac{5}{4}+10}$

$y_{cm}=\dfrac{7}{3}\ m$

Hence, The distance of center of mass  is  $\dfrac{7}{3}$

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