Lamina with removed triangle/rectangle/square

\includegraphics{figure_3} A uniform lamina is in the form of a triangle \(ABC\), with \(AC = 8a\), \(BC = 6a\) and angle \(ACB = 90°\). The point \(D\) on \(AC\) is such that \(AD = 3a\). The point \(E\) on \(CB\) is such that \(CE = x\) (see diagram). The triangle \(CDE\) is removed from the lamina.

1. Find, in terms of \(a\) and \(x\), the distance of the centre of mass of the remaining object \(ADEB\) from \(AC\). [4]

The object \(ADEB\) is on the point of toppling about the point \(E\) when the object is in the vertical plane with its edge \(EB\) on a smooth horizontal surface.

1. Find \(x\) in terms of \(a\). [3]

\includegraphics{figure_5} A uniform lamina is in the form of a triangle \(OBC\), with \(OC = 18a\), \(OB = 24a\) and angle \(COB = 90°\). The point \(A\) on \(OB\) is such that \(OA = x\) (see diagram). The triangle \(OAC\) is removed from the lamina.

1. Find, in terms of \(a\) and \(x\), the distance of the centre of mass of the remaining object \(ABC\) from \(OC\). [3]

The object \(ABC\) is suspended from \(C\). In its equilibrium position, the side \(AB\) makes an angle \(\theta\) with the vertical, where \(\tan\theta = \frac{8}{5}\).

1. Find \(x\) in terms of \(a\). [4]

\includegraphics{figure_4} A uniform lamina \(AECF\) is formed by removing two identical triangles \(BCE\) and \(CDF\) from a square lamina \(ABCD\). The square has side \(3a\) and \(EB = DF = h\) (see diagram).

1. Find the distance of the centre of mass of the lamina \(AECF\) from \(AD\) and from \(AB\), giving your answers in terms of \(a\) and \(h\). [5]

The lamina \(AECF\) is placed vertically on its edge \(AE\) on a horizontal plane.

1. Find, in terms of \(a\), the set of values of \(h\) for which the lamina remains in equilibrium. [3]

\includegraphics{figure_2} A uniform lamina is in the form of a triangle \(ABC\) in which angle \(B\) is a right angle, \(AB = 9a\) and \(BC = 6a\). The point \(D\) is on \(BC\) such that \(BD = x\) (see diagram). The region \(ABD\) is removed from the lamina. The resulting shape \(ADC\) is placed with the edge \(DC\) on a horizontal surface and the plane \(ADC\) is vertical. Find the set of values of \(x\), in terms of \(a\), for which the shape is in equilibrium. [6]

\includegraphics{figure_3} A uniform lamina is in the form of an isosceles triangle \(ABC\) in which \(AC = 2a\) and angle \(ABC = 90°\). The point \(D\) on \(AB\) is such that the ratio \(DB : AB = 1 : k\). The point \(E\) on \(CB\) is such that \(DE\) is parallel to \(AC\). The triangle \(DBE\) is removed from the lamina (see diagram).

1. Find, in terms of \(k\), the distance of the centre of mass of the remaining lamina \(ADEC\) from the midpoint of \(AC\). [4]

When the lamina \(ADEC\) is freely suspended from the vertex \(A\), the edge \(AC\) makes an angle \(\theta\) with the downward vertical, where \(\tan \theta = \frac{2}{15}\).

1. Find the value of \(k\). [3]

\includegraphics{figure_1} A uniform lamina \(ABCD\) is formed by removing the isosceles triangle \(ADC\) of height \(h\) metres, where \(h < 2\sqrt{3}\), from a uniform lamina \(ABC\) in the shape of an equilateral triangle of side 4 m, as shown in Figure 1. The centre of mass of \(ABCD\) is at \(D\).

1. Show that \(h = \sqrt{3}\) [7]

The weight of the lamina \(ABCD\) is \(W\) newtons. The lamina is freely suspended from \(A\). A horizontal force of magnitude \(F\) newtons is applied at \(B\) so that the lamina is in equilibrium with \(AB\) vertical. The horizontal force acts in the vertical plane containing the lamina.

1. Find \(F\) in terms of \(W\). [4]

\includegraphics{figure_1} Figure 1 shows a template made by removing a square \(WXYZ\) from a uniform triangular lamina \(ABC\). The lamina is isosceles with \(CA = CB\) and \(AB = 12a\). The midpoint of \(AB\) is \(N\) and \(NC = 8a\). The centre \(O\) of the square lies on \(NC\) and \(ON = 2a\). The sides \(WX\) and \(ZY\) are parallel to \(AB\) and \(WZ = 2a\). The centre of mass of the template is at \(G\).

1. Show that \(NG = \frac{8}{7}a\). [7]

The template has mass \(M\). A small metal stud of mass \(kM\) is attached to the template at \(C\). The centre of mass of the combined template and stud lies on \(YZ\). By modelling the stud as a particle,

1. calculate the value of \(k\). [4]

\includegraphics{figure_1} Figure 1 shows a template made by removing a square \(WXYZ\) from a uniform triangular lamina \(ABC\). The lamina is isosceles with \(CA = CB\) and \(AB = 12a\). The mid-point of \(AB\) is \(N\) and \(NC = 8a\). The centre \(O\) of the square lies on \(NC\) and \(ON = 2a\). The sides \(WX\) and \(ZY\) are parallel to \(AB\) and \(WZ = 2a\). The centre of mass of the template is at \(G\).

1. Show that \(NG = \frac{30}{11}a\). [7]

The template has mass \(M\). A small metal stud of mass \(kM\) is attached to the template at \(C\). The centre of mass of the combined template and stud lies on \(YZ\). By modelling the stud as a particle,

1. calculate the value of \(k\). [4]

\includegraphics{figure_1} A uniform lamina \(ABCD\) is made by taking a uniform sheet of metal in the form of a rectangle \(ABED\), with \(AB = 3a\) and \(AD = 2a\), and removing the triangle \(BCE\), where \(C\) lies on \(DE\) and \(CE = a\), as shown in Fig. 1.

1. Find the distance of the centre of mass of the lamina from \(AD\). [5]

The lamina has mass \(M\). A particle of mass \(m\) is attached to the lamina at \(B\). When the loaded lamina is freely suspended from the mid-point of \(AB\), it hangs in equilibrium with \(AB\) horizontal.

1. Find \(m\) in terms of \(M\). [4]

\includegraphics{figure_3} Figure 3 shows a uniform equilateral triangular lamina \(PRT\) with sides of length \(2a\).

1. Using calculus, prove that the centre of mass of \(PRT\) is at a distance \(\frac{2\sqrt{3}}{3}a\) from \(R\). [6]

\includegraphics{figure_4} The circular sector \(PQU\), of radius \(a\) and centre \(P\), and the circular sector \(TUS\), of radius \(a\) and centre \(T\), are removed from \(PRT\) to form the uniform lamina \(QRSU\) shown in Figure 4.

1. Show that the distance of the centre of mass of \(QRSU\) from \(U\) is \(\frac{2a}{3\sqrt{3} - \pi}\). [6]

\includegraphics{figure_4} The diagram shows a uniform lamina \(ABCDE\) formed by removing a symmetrical triangular section from a rectangular sheet of metal measuring 30 cm by 25 cm.

1. Find the distance of the centre of mass of the lamina from \(ED\). [4 marks]

The lamina has mass \(m\). A particle \(P\) is attached to the lamina at \(B\). The lamina is then suspended freely from \(A\) and hangs in equilibrium with \(AD\) vertical.

1. Find, in terms of \(m\), the mass of \(P\). [5 marks]