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In this lesson, we will learn how to find the position of the center of gravity (mass) of a standard uniform plane lamina.
Q1:
A uniform triangular lamina π΄ π΅ πΆ is right-angled at π΅ , π΅ πΆ = 1 7 c m , π΄ π΅ = 1 7 c m , and π , π , and π are the midpoints of π΄ π΅ , π΅ πΆ , and πΆ π΄ respectively. Triangle πΆ π π was cut off and then affixed to the lamina above triangle π π΅ π . The body was freely suspended from point π΅ . Find the tangent of the angle that π΅ πΆ makes to the vertical, t a n π , when the body is hanging in its equilibrium position.
Q2:
A uniform rectangular lamina has a length of 63 cm and a width of 59 cm. It is divided into three equal sized rectangles along its length, the last of these rectangles has been folded over so that it lies flat on the middle rectangle as show in the figure. Find the coordinates of the center of gravity of the lamina in this form.
Q3:
Where does the centre of gravity of a uniform circular disc lie?
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