Lamina with attached triangle

4. \begin{figure}[h] \end{figure} The uniform lamina \(A B C\) is an isosceles triangle with \(A B = B C , A C = 6 a\) and the distance from \(B\) to \(A C\) is \(3 a\). The uniform lamina \(M N C\) is an isosceles triangle with \(M N = N C\) and \(M C = 3 a\). Triangles \(A B C\) and \(M N C\) are similar and are made of the same material. The lamina \(L\) is formed by fixing triangle \(M N C\) on top of triangle \(A B C\), as shown in Figure 2.

1. Show that the distance of the centre of mass of \(L\) from \(A C\) is \(\frac { 9 } { 10 } a\) The lamina \(L\) is freely suspended from \(B\) and hangs in equilibrium.
2. Find, to the nearest degree, the size of the angle between \(A B\) and the downward vertical.

7. \begin{figure}[h] \end{figure} The template shown in Figure 3 is formed by joining together three separate laminas. All three laminas lie in the same plane.

• PQUV is a uniform square lamina with sides of length \(3 a\)
• URST is a uniform square lamina with sides of length \(6 a\)
• \(Q R U\) is a uniform triangular lamina with \(U Q = 3 a , U R = 6 a\) and angle \(Q U R = 90 ^ { \circ }\)

The mass per unit area of \(P Q U V\) is \(k\), where \(k\) is a constant.
The mass per unit area of URST is \(k\).
The mass per unit area of \(Q R U\) is \(2 k\).
The distance of the centre of mass of the template from \(Q T\) is \(d\).

1. Show that \(d = \frac { 29 } { 14 } a\) The template is freely suspended from the point \(Q\) and hangs in equilibrium with \(Q R\) at \(\theta ^ { \circ }\) to the downward vertical.
2. Find the value of \(\theta\)

5. \begin{figure}[h] \end{figure} A uniform lamina \(A B C D\) is made by joining a uniform triangular lamina \(A B D\) to a uniform semi-circular lamina \(D B C\), of the same material, along the edge \(B D\), as shown in Figure 2. Triangle \(A B D\) is right-angled at \(D\) and \(A D = 18 \mathrm {~cm}\). The semi-circle has diameter \(B D\) and \(B D = 12 \mathrm {~cm}\).

1. Show that, to 3 significant figures, the distance of the centre of mass of the lamina \(A B C D\) from \(A D\) is 4.69 cm . Given that the centre of mass of a uniform semicircular lamina, radius \(r\), is at a distance \(\frac { 4 r } { 3 \pi }\) from the centre of the bounding diameter,
2. find, in cm to 3 significant figures, the distance of the centre of mass of the lamina \(A B C D\) from \(B D\). The lamina is freely suspended from \(B\) and hangs in equilibrium.
3. Find, to the nearest degree, the angle which \(B D\) makes with the vertical.

3. \begin{figure}[h] \end{figure} Figure 1 shows a decoration which is made by cutting the shape of a simple tree from a sheet of uniform card. The decoration consists of a triangle \(A B C\) and a rectangle \(P Q R S\). The points \(P\) and \(S\) lie on \(B C\) and \(M\) is the mid-point of both \(B C\) and \(P S\). The triangle \(A B C\) is isosceles with \(A B = A C , B C = 4 \mathrm {~cm} , A M = 6 \mathrm {~cm} , P S = 2 \mathrm {~cm}\) and \(P Q = 3 \mathrm {~cm}\).

1. Find the distance of the centre of mass of the decoration from \(B C\). The decoration is suspended from \(Q\) and hangs freely.
2. Find, in degrees to one decimal place, the angle between \(P Q\) and the vertical.

5. \begin{figure}[h] \end{figure} A shop sign \(A B C D E F G\) is modelled as a uniform lamina, as illustrated in Figure 2. \(A B C D\) is a rectangle with \(B C = 120 \mathrm {~cm}\) and \(D C = 90 \mathrm {~cm}\). The shape \(E F G\) is an isosceles triangle with \(E G = 60 \mathrm {~cm}\) and height 60 cm . The mid-point of \(A D\) and the mid-point of \(E G\) coincide.

1. Find the distance of the centre of mass of the sign from the side \(A D\). The sign is freely suspended from \(A\) and hangs at rest.
2. Find the size of the angle between \(A B\) and the vertical.

3. [The centre of mass of a semicircular lamina of radius \(r\) is \(\frac { 4 r } { 3 \pi }\) from the centre.] \begin{figure}[h] \end{figure} Figure 2 shows the uniform lamina \(A B C D E\), such that \(A B D E\) is a square with sides of length \(2 a\) and \(B C D\) is a semicircle with diameter \(B D\).

1. Show that the distance of the centre of mass of the lamina from \(B D\) is \(\frac { 20 a } { 3 ( 8 + \pi ) }\). The lamina is freely suspended from \(D\) and hangs in equilibrium.
2. Find, to the nearest degree, the angle that \(D E\) makes with the downward vertical.

7. \includegraphics[max width=\textwidth, alt={}, center]{0d3d35b1-e3c5-47ac-b05e-78cdf1eb3083-4_360_472_1105_815} A uniform plane lamina \(A B C D E\) is formed by joining a uniform square \(A B D E\) with a uniform triangular lamina \(B C D\), of the same material, along the side \(B D\), as shown in Fig. 2. The lengths \(A B , B C\) and \(C D\) are \(18 \mathrm {~cm} , 15 \mathrm {~cm}\) and 15 cm respectively.

1. Find the distance of the centre of mass of the lamina from \(A E\). The lamina is freely suspended from \(B\) and hangs in equilibrium.
2. Find, in degrees to one decimal place, the angle which \(B D\) makes with the vertical.

6 \includegraphics[max width=\textwidth, alt={}, center]{1fbb3693-0beb-47c8-800f-50041f105699-3_540_878_989_632} A uniform lamina \(A B C D E\) of weight 30 N consists of a rectangle and a right-angled triangle. The dimensions are as shown in the diagram.

1. Taking \(x\) - and \(y\)-axes along \(A E\) and \(A B\) respectively, find the coordinates of the centre of mass of the lamina. The lamina is freely suspended from a hinge at \(B\).
2. Calculate the angle that \(A B\) makes with the vertical. The lamina is now held in a position such that \(B D\) is horizontal. This is achieved by means of a string attached to \(D\) and to a fixed point 15 cm directly above the hinge at \(B\).
3. Calculate the tension in the string.

5 \includegraphics[max width=\textwidth, alt={}, center]{d6d87705-be4b-407d-b699-69fb441d88a7-3_657_549_1219_799} A uniform lamina \(A B C D E\) consists of a square and an isosceles triangle. The square has sides of 18 cm and \(B C = C D = 15 \mathrm {~cm}\) (see diagram).

1. Taking \(x\) - and \(y\)-axes along \(A E\) and \(A B\) respectively, find the coordinates of the centre of mass of the lamina.
2. The lamina is freely suspended from \(B\). Calculate the angle that \(B D\) makes with the vertical. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d6d87705-be4b-407d-b699-69fb441d88a7-4_441_1355_265_394} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} A light inextensible string of length 1 m passes through a small smooth hole \(A\) in a fixed smooth horizontal plane. One end of the string is attached to a particle \(P\), of mass 0.5 kg , which hangs in equilibrium below the plane. The other end of the string is attached to a particle \(Q\), of mass 0.3 kg , which rotates with constant angular speed in a circle of radius 0.2 m on the surface of the plane (see Fig. 1).

3 A uniform circular lamina, of radius 4 cm and mass 0.4 kg , has a centre \(O\), and \(A B\) is a diameter. To create a medal, a smaller uniform circular lamina, of radius 2 cm and mass 0.1 kg , is attached so that the centre of the smaller lamina is at the point \(A\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-06_671_878_513_598}

1. Explain why the centre of mass of the medal is on the line \(A B\).
2. Find the distance of the centre of mass of the medal from the point \(B\). \includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-06_1259_1705_1448_155}
   \includegraphics[max width=\textwidth, alt={}]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-07_2484_1709_223_153}

\includegraphics{figure_4} The diagram shows a uniform lamina \(ABCD\) with \(AB = 0.75 \text{ m}\), \(AD = 0.6 \text{ m}\) and \(BC = 0.9 \text{ m}\). Angle \(BAD =\) angle \(ABC = 90°\).

1. Show that the distance of the centre of mass of the lamina from \(AB\) is \(0.38 \text{ m}\), and find the distance of the centre of mass from \(BC\). [5]

The lamina is freely suspended at \(B\) and hangs in equilibrium.

1. Find the angle between \(BC\) and the vertical. [2]

\includegraphics{figure_1} \(ABCD\) is a uniform lamina with \(AB = 1.8\) m, \(AD = DC = 0.9\) m, and \(AD\) perpendicular to \(AB\) and \(DC\) (see diagram).

1. Find the distance of the centre of mass of the lamina from \(AB\) and the distance from \(AD\). [4]

The lamina is freely suspended at \(A\) and hangs in equilibrium.

1. Calculate the angle between \(AB\) and the vertical. [2]

\includegraphics{figure_6} A uniform lamina \(OABCD\) consists of a semicircle \(BCD\) with centre \(O\) and radius \(0.6\) m and an isosceles triangle \(OAB\), joined along \(OB\) (see diagram). The triangle has area \(0.36\) m\(^2\) and \(AB = AO\).

1. Show that the centre of mass of the lamina lies on \(OB\). [4]
2. Calculate the distance of the centre of mass of the lamina from \(O\). [4]

\includegraphics{figure_1} A uniform lamina \(ABCD\) consists of two isosceles triangles \(ABD\) and \(BCD\). The diagonals of \(ABCD\) meet at the point \(O\). The length of \(AO\) is \(3a\), the length of \(OC\) is \(6a\) and the length of \(BD\) is \(16a\) (see diagram). Find the distance of the centre of mass of the lamina from \(DB\). [3]

\includegraphics{figure_2} Figure 2 shows a uniform lamina \(ABCDE\) such that \(ABDE\) is a rectangle, \(BC = CD\), \(AB = 8a\) and \(AE = 6a\). The point \(X\) is the mid-point of \(BD\) and \(XC = 4a\). The centre of mass of the lamina is at \(G\).

1. Show that \(GX = \frac{14}{15}a\). [6]

The mass of the lamina is \(M\). A particle of mass \(\lambda M\) is attached to the lamina at \(C\). The lamina is suspended from \(B\) and hangs freely under gravity with \(AB\) horizontal.

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

\includegraphics{figure_1} The trapezium \(ABCD\) is a uniform lamina with \(AB = 4\) m and \(BC = CD = DA = 2\) m, as shown in Figure 1.

1. Show that the centre of mass of the lamina is \(\frac{4\sqrt{3}}{9}\) m from \(AB\). [5]

The lamina is freely suspended from \(D\) and hangs in equilibrium.

1. Find the angle between \(DC\) and the vertical through \(D\). [5]

\includegraphics{figure_1} Figure 1 shows a uniform lamina \(ABCDE\) such that \(ABDE\) is a rectangle, \(BC = CD\), \(AB = 4a\) and \(AE = 2a\). The point \(F\) is the midpoint of \(BD\) and \(FC = a\).

1. Find, in terms of \(a\), the distance of the centre of mass of the lamina from \(AE\). [4]

The lamina is freely suspended from \(A\) and hangs in equilibrium.

1. Find the angle between \(AB\) and the downward vertical. [3]

A uniform lamina is in the form of a trapezium \(ABCD\), as shown. \(AB\) and \(DC\) are perpendicular to \(BC\). \(AB = 17\) cm, \(BC = 21\) cm and \(CD = 8\) cm. \includegraphics{figure_7}

1. Find the distances of the centre of mass of the lamina from
   1. \(AB\),
   2. \(BC\). [8 marks]

The lamina is freely suspended from \(C\) and rests in equilibrium.

1. Find the angle between \(CD\) and the vertical. [3 marks]

You are given that the centre of mass, G, of a uniform lamina in the shape of an isosceles triangle lies on its axis of symmetry in the position shown in Fig. 4.1. \includegraphics{figure_4_1} Fig. 4.2 shows the cross-section OABCD of a prism made from uniform material. OAB is an isosceles triangle, where OA = AB, and OBCD is a rectangle. The distance OD is \(h\) cm, where \(h\) can take various positive values. All coordinates refer to the axes Ox and Oy shown. The units of the axes are centimetres. \includegraphics{figure_4_2}

1. Write down the coordinates of the centre of mass of the triangle OAB. [1]
2. Show that the centre of mass of the region OABCD is \(\left(\frac{12-h^2}{2(h+3)}, 2.5\right)\). [6]

The \(x\)-axis is horizontal. The prism is placed on a horizontal plane in the position shown in Fig. 4.2.

1. Find the values of \(h\) for which the prism would topple. [3]

The following questions refer to the case where \(h = 3\) with the prism held in the position shown in Fig. 4.2. The cross-section OABCD contains the centre of mass of the prism. The weight of the prism is 15 N. You should assume that the prism does not slide.

1. Suppose that the prism is held in this position by a vertical force applied at A. Given that the prism is on the point of tipping clockwise, calculate the magnitude of this force. [3]
2. Suppose instead that the prism is held in this position by a force in the plane of the cross-section OABCD, applied at 30° below the horizontal at C, as shown in Fig. 4.3. Given that the prism is on the point of tipping anti-clockwise, calculate the magnitude of this force. [4]

\includegraphics{figure_4_3}

\includegraphics{figure_4} Fig. 4 shows a uniform lamina ABCDE such that ABDE is a rectangle and BCD is an isosceles triangle. AB = 5a, AE = 4a and BC = CD. The point F is the midpoint of BD and FC = a.

1. Find, in terms of \(a\), the exact distance of the centre of mass of the lamina from AE. [4]

The lamina is freely suspended from B and hangs in equilibrium.

1. Find the angle between AB and the downward vertical. [2]