Suspended lamina equilibrium angle

1. \hspace{0pt} [The centre of mass of a semicircular arc of radius \(r\) is \(\frac { 2 r } { \pi }\) from the centre.]

\begin{figure}[h] \end{figure} Uniform wire is used to form the framework shown in Figure 2.
In the framework,

• \(A B C\) is straight and has length \(25 a\)
• \(A D E\) is straight and has length \(24 a\)
• \(A B D\) is a semicircular arc of radius \(7 a\)
• \(E C = 7 a\)
• angle \(A E C = 90 ^ { \circ }\)
• the points \(A , B , C , D\) and \(E\) all lie in the same plane

The distance of the centre of mass of the framework from \(A E\) is \(d\).

1. Show that \(d = \frac { 53 } { 2 ( 7 + \pi ) } a\) The framework is freely suspended from \(A\) and hangs in equilibrium with \(A C\) at angle \(\alpha ^ { \circ }\) to the downward vertical.
2. Find the value of \(\alpha\).

3. \begin{figure}[h] \end{figure} The uniform lamina \(A B D E\) is in the shape of a rectangle with \(A B = 8 a\) and \(B D = 6 a\). The triangle \(B C D\) is isosceles and has base \(6 a\) and perpendicular height \(6 a\). The template \(A B C D E\), shown shaded in Figure 1, is formed by removing the triangular lamina \(B C D\) from the lamina \(A B D E\).

1. Show that the centre of mass of the template is \(\frac { 14 } { 5 } a\) from \(A E\). The template is freely suspended from \(A\) and hangs in equilibrium with \(A B\) at an angle of \(\theta ^ { \circ }\) to the downward vertical.
2. Find the value of \(\theta\), giving your answer to the nearest whole number.

5. \begin{figure}[h] \end{figure} The uniform lamina \(A B C\) is in the shape of an equilateral triangle with sides of length \(4 a\). The midpoint of \(B C\) is \(D\). The point \(E\) lies on \(A D\) with \(D E = \frac { 3 a } { 2 }\). A circular hole, with centre \(E\) and radius \(a\), is made in the lamina \(A B C\) to form the lamina \(L\), shown shaded in Figure 2.

1. Find the distance of the centre of mass of \(L\) from \(D\). The lamina \(L\) is freely suspended from the point \(B\) and hangs in equilibrium.
2. Find, to the nearest degree, the size of the acute angle between \(A D\) and the downward vertical.

7. In this question you may use, without proof, the formula for the centre of mass of a uniform sector of a circle, as given in the formulae book. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E\), shown shaded in Figure 3, is formed by joining a rectangle to a sector of a circle.

• The rectangle \(A B C E\) has \(A B = E C = a\) and \(A E = B C = d\)
• The sector \(C D E\) has centre \(C\) and radius \(a\)
• Angle \(E C D = \frac { \pi } { 3 }\) radians

The centre of mass of the lamina lies on EC.

1. Show that \(a = \sqrt { 3 } d\) The lamina is freely suspended from \(B\) and hangs in equilibrium with \(B C\) at an angle \(\beta\) radians to the downward vertical.
2. Find the value of \(\beta\)

3. \begin{figure}[h] \end{figure} A uniform lamina \(A B C D E F\) is formed by taking a uniform sheet of card in the form of a square \(A X E F\), of side \(2 a\), and removing the square \(B X D C\) of side \(a\), where \(B\) and \(D\) are the mid-points of \(A X\) and \(X E\) respectively, as shown in Figure 1.

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

4. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E F\) is a regular hexagon with centre \(O\) and sides of length 2 m , as shown in Figure 1. \begin{figure}[h] \end{figure} The triangles \(O A F\) and \(O E F\) are removed to form the uniform lamina \(O A B C D E\), shown in Figure 2.

1. Find the distance of the centre of mass of \(O A B C D E\) from \(O\). The lamina \(O A B C D E\) is freely suspended from \(E\) and hangs in equilibrium.
2. Find the size of the angle between \(E O\) and the downward vertical.

4. \begin{figure}[h] \end{figure} Figure 2 shows a lamina \(L\). It is formed by removing a square \(P Q R S\) from a uniform triangle \(A B C\). The triangle \(A B C\) is isosceles with \(A C = B C\) and \(A B = 12 \mathrm {~cm}\). The midpoint of \(A B\) is \(D\) and \(D C = 8 \mathrm {~cm}\). The vertices \(P\) and \(Q\) of the square lie on \(A B\) and \(P Q = 4 \mathrm {~cm}\). The centre of the square is \(O\). The centre of mass of \(L\) is at \(G\).

1. Find the distance of \(G\) from \(A B\). When \(L\) is freely suspended from \(A\) and hangs in equilibrium, the line \(A B\) is inclined at \(25 ^ { \circ }\) to the vertical.
2. Find the distance of \(O\) from \(D C\).

3. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E F\), shown shaded in Figure 1, is symmetrical about the line through \(B\) and \(E\). It is formed by removing the isosceles triangle \(F E D\), of height \(6 a\) and base \(8 a\), from the isosceles triangle \(A B C\) of height \(9 a\) and base \(12 a\).

1. Find, in terms of \(a\), the distance of the centre of mass of the lamina from \(A C\). The lamina is freely suspended from \(A\) and hangs in equilibrium.
2. Find, to the nearest degree, the size of the angle between \(A B\) and the downward vertical.

3. \begin{figure}[h] \end{figure} A uniform plane lamina is in the shape of an isosceles triangle \(A B C\), where \(A B = A C\). The mid-point of \(B C\) is \(M , A M = 30 \mathrm {~cm}\) and \(B M = 40 \mathrm {~cm}\). The mid-points of \(A C\) and \(A B\) are \(D\) and \(E\) respectively. The triangular portion \(A D E\) is removed leaving a uniform plane lamina \(B C D E\) as shown in Fig. 2.

1. Show that the centre of mass of the lamina \(B C D E\) is \(6 \frac { 2 } { 3 } \mathrm {~cm}\) from \(B C\).
   (6 marks)
   The lamina \(B C D E\) is freely suspended from \(D\) and hangs in equilibrium.
2. Find, in degrees to one decimal place, the angle which \(D E\) makes with the vertical.
   (3 marks)

4. \begin{figure}[h] \end{figure} The uniform lamina \(O B C\) is one quarter of a circular disc with centre \(O\) and radius 4 m . The points \(A\) and \(D\), on \(O B\) and \(O C\) respectively, are 3 m from \(O\). The uniform lamina \(A B C D\), shown shaded in Figure 1, is formed by removing the triangle \(O A D\) from \(O B C\). Given that the centre of mass of one quarter of a uniform circular disc of radius \(r\) is at a distance \(\frac { 4 \sqrt { 2 } } { 3 \pi } r\) from the centre of the disc,

1. find the distance of the centre of mass of the lamina \(A B C D\) from \(A D\). The lamina is freely suspended from \(D\) and hangs in equilibrium.
2. Find, to the nearest degree, the angle between \(D C\) and the downward vertical.
   \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{deb9e495-3bfb-4a46-9ee7-3eb421c33499-09_915_1269_118_356} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure}

8

1. A uniform semicircular lamina has radius 4 cm . Show that the distance from its centre to its centre of mass is 1.70 cm , correct to 3 significant figures.
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{982647bd-8514-40cf-b4ee-674f51df32c5-4_429_947_405_640} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} A model bridge is made from a uniform rectangular board, \(A B C D\), with a semicircular section, \(E F G\), removed. \(O\) is the mid-point of \(E G\). \(A B = 8 \mathrm {~cm} , B C = 20 \mathrm {~cm} , A O = 12 \mathrm {~cm}\) and the radius of the semicircle is 4 cm (see Fig. 1).
   1. Show that the distance from \(A B\) to the centre of mass of the model is 9.63 cm , correct to 3 significant figures.
   2. Calculate the distance from \(A D\) to the centre of mass of the model.
   3. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{982647bd-8514-40cf-b4ee-674f51df32c5-4_572_945_1416_641} \captionsetup{labelformat=empty} \caption{Fig. 2}
      \end{figure} The model bridge is smoothly pivoted at \(A\) and is supported in equilibrium by a vertical wire attached to \(D\). The weight of the model is 15 N and \(A D\) makes an angle of \(10 ^ { \circ }\) with the horizontal (see Fig. 2). Calculate the tension in the wire.

3 A uniform lamina \(A B C D E\) consists of a square \(A C D E\) and an equilateral triangle \(A B C\) which are joined along their common edge \(A C\) to form a pentagon whose sides are each 8 cm in length.

1. Calculate the distance of the centre of mass of the lamina from \(A C\).
2. The lamina is freely suspended from \(A\) and hangs in equilibrium. Calculate the angle that \(A C\) makes with the vertical.

3 \includegraphics[max width=\textwidth, alt={}, center]{5bfd0285-71cb-4dcb-8545-a379653f9a3e-2_778_579_1304_744} A uniform lamina \(A B C D E\) consists of a rectangle \(A B D E\) and an isosceles triangle \(B C D\) joined along their common edge. \(A B = D E = 8 \mathrm {~cm} , A E = B D = 10 \mathrm {~cm}\) and \(B C = C D = 13 \mathrm {~cm}\) (see diagram).

1. Find the distance of the centre of mass of the lamina from \(A E\).
2. The lamina is freely suspended from \(B\) and hangs in equilibrium. Calculate the angle that \(B D\) makes with the vertical.

5. \begin{figure}[h] \end{figure} Figure 2 shows a uniform plane lamina \(A B C D E G\) in the shape of a letter ' \(L\) ' consisting of a rectangle \(A B F G\) joined to another rectangle \(C D E F\). The sides \(A B\) and \(D E\) are both 8 cm long and the sides \(E G\) and \(G A\) are of length 24 cm and 32 cm respectively.

1. Show that the centre of mass of the lamina lies on the line \(B F\).
2. Find the distance of the centre of mass from the line \(A B\). The uniform lamina in Figure 2 is a model of the letter ' \(L\) ' in a sign above a shop. The letter is normally suspended from a wall at \(A\) and \(B\) so that \(A B\) is horizontal but the fixing at \(B\) has broken and the letter hangs in equilibrium from the point \(A\).
3. Find, in degrees to one decimal place, the acute angle \(A G\) makes with the vertical.

7. \begin{figure}[h] \end{figure} Figure 2 shows a uniform lamina \(A B C D\) formed by removing an isosceles triangle \(B C D\) from an equilateral triangle \(A B D\) of side \(2 d\). The point \(C\) is the centroid of triangle \(A B D\).

1. Find the area of triangle \(B C D\) in terms of \(d\).
2. Show that the distance of the centre of mass of the lamina from \(B D\) is \(\frac { 4 } { 9 } \sqrt { 3 } d\).
   (8 marks)
   The lamina is freely suspended from the point \(B\) and hangs at rest.
3. Find in degrees, correct to 1 decimal place, the acute angle that the side \(A B\) makes with the vertical.

7. \begin{figure}[h] \end{figure} Figure 3 shows a hotel 'key' consisting of a rectangle \(O A B D\), where \(O A = 8 \mathrm {~cm}\) and \(O D = 4 \mathrm {~cm}\), joined to a semicircle whose diameter \(B C\) is 4 cm long. The thickness of the key is negligible and the same material is used throughout. The key is modelled as a uniform lamina.
Using this model,

1. find, correct to 3 significant figures, the distance of the centre of mass from
   1. OD ,
   2. \(O A\). A small circular hole of negligible diameter is made at the mid-point of \(B C\) so that the key can be hung on a smooth peg. When the key is freely suspended from the peg,
2. find, correct to 3 significant figures, the acute angle made by \(O A\) with the vertical.

4. \begin{figure}[h] \end{figure} Figure 2 shows an earring consisting of a uniform wire \(A B C D\) of length \(6 a\) bent to form right angles at \(B\) and \(C\) such that \(A B\) and \(C D\) are of length \(2 a\) and \(a\) respectively.

1. Find, in terms of \(a\), the distance of the centre of mass from
   1. \(\quad A B\),
   2. \(B C\). The earring is to be worn such that it hangs in equilibrium suspended from the point \(A\).
2. Find, to the nearest degree, the angle made by \(A B\) with the downward vertical.
   (4 marks)

4 A uniform lamina ABDE is in the shape of an equilateral triangle ABC of side 12 cm from which an equilateral triangle of side 6 cm has been removed from corner \(C\). The lamina is situated on coordinate axes as shown in Fig. 4. \begin{figure}[h] \end{figure}

1. Explain why angle \(\mathrm { BDA } = 90 ^ { \circ }\).
2. Find the coordinates of the centre of mass of the lamina ABDE . The lamina ABDE is now freely suspended from D and hangs in equilibrium.
3. Calculate the angle DE makes with the downward vertical.

6 A uniform lamina OABC is in the shape of a trapezium where O is the origin of the coordinate system in which the points \(A , B\) and \(C\) have coordinates \(( 12,0 ) , ( 12 + p , q )\) and \(( 0 , q )\) respectively. \includegraphics[max width=\textwidth, alt={}, center]{a96a0ebe-8f4f-4d79-9d11-9d348ef72314-7_536_917_349_239}

1. Determine, in terms of \(p\) and \(q\), the coordinates of the centre of mass of OABC . The point D has coordinates \(( 7.6 , q )\). When OABC is suspended from D , the lamina hangs in equilibrium with BC horizontal.
2. Determine the value of \(p\). When OABC is suspended from C, the lamina hangs in equilibrium with BC at an angle of \(35 ^ { \circ }\) to the downward vertical.
3. Determine the value of \(q\), giving your answer correct to \(\mathbf { 3 }\) significant figures.

5 Fig. 5 shows the curve with equation \(y = - x ^ { 2 } + 4 x + 2\).
The curve intersects the \(x\)-axis at P and Q . The region bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 4\) is occupied by a uniform lamina L . The horizontal base of L is OA , where A is the point \(( 4,0 )\). \begin{figure}[h] \end{figure}

1. 1. Explain why the centre of mass of L lies on the line \(x = 2\).
   2. In this question you must show detailed reasoning. Find the \(y\)-coordinate of the centre of mass of \(L\).
2. L is freely suspended from A . Find the angle AO makes with the vertical. The region bounded by the curve and the \(x\)-axis is now occupied by a uniform lamina M . The horizontal base of M is PQ.
3. Explain how the position of the centre of mass of M differs from the position of the centre of mass of \(L\).

4. The diagram shows a uniform lamina consisting of a rectangular section \(G P Q E\) with a semi-circular section EFG of radius 4 cm . Quadrants \(A P B\) and \(C Q D\) each with radius 2 cm are removed. Dimensions in cm are as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{3efc4ef6-8a80-4267-8e95-733200e875c5-3_758_604_497_651}

1. Write down the distance of the centre of mass of the lamina \(A B C D E F G\) from \(A G\).
2. Determine the distance of the centre of mass of the lamina \(A B C D E F G\) from \(B C\).
3. The lamina \(A B C D E F G\) is suspended freely from the point \(E\) and hangs in equilibrium. Calculate the angle EG makes with the vertical.

3. \begin{figure}[h] \end{figure} The lamina \(L\), shown in Figure 2, consists of a uniform square lamina \(A B D F\) and two uniform triangular laminas \(B D C\) and \(F D E\). The square has sides of length \(2 a\). The two triangles are identical. The straight lines \(B D E\) and \(F D C\) are perpendicular with \(B D = D F = 2 a\) and \(D C = D E = a\).
The mass per unit of area of the square is \(M\).
The mass per unit area of each triangle is \(3 M\).
The centre of mass of \(L\) is at the point \(G\).

1. Without doing any calculations, explain why \(G\) lies on \(A D\).
2. Show that the distance of \(G\) from \(D\) is \(\frac { \sqrt { 2 } } { 2 } a\) The lamina \(L\) is freely suspended from \(B\) and hangs in equilibrium.
3. Find the size of the angle between \(B E\) and the downward vertical.

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4. \begin{figure}[h] \end{figure} The uniform triangular lamina \(A B C\) has \(A B\) perpendicular to \(A C\), \(A B = 9 a\) and \(A C = 6 a\). The point \(D\) on \(A B\) is such that \(A D = a\). The rectangle \(D E F G\), with \(D E = 2 a\) and \(E F = 3 a\), is removed from the lamina to form the template shown shaded in Figure 3. The distance of the centre of mass of the template from \(A C\) is \(d\).

1. Show that \(d = \frac { 23 } { 7 } a\) The template is freely suspended from \(A\) and hangs in equilibrium with \(A B\) at an angle \(\theta ^ { \circ }\) to the downward vertical through \(A\).
2. Find the value of \(\theta\) A new piece, of exactly the same size and shape as the template, is cut from a lamina of a different uniform material. The template and the new piece are joined together to form the model shown in Figure 4. Both parts of the model lie in the same plane. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{fd8bc7b5-adee-4d67-b15d-571255b00b83-12_369_1185_1667_440} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} The weight of \(C P Q R S T A\) is \(W\) The weight of \(A D G F E B C\) is \(4 W\) The model is freely suspended from \(A\).
   A horizontal force of magnitude \(X\), acting in the same vertical plane as the model, is now applied to the model at \(T\) so that \(A C\) is vertical, as shown in Figure 4.
3. Find \(X\) in terms of \(W\).

1. \begin{figure}[h] \end{figure} A letter P from a shop sign is modelled as a uniform plane lamina which consists of a rectangular lamina, \(O A B D E\), joined to a semicircular lamina, \(B C D\), along its diameter \(B D\). $$O A = E D = a , A B = 2 a , O E = 4 a \text {, and the diameter } B D = 2 a \text {, as shown in Figure } 1 .$$ Using the model,

1. find, in terms of \(\pi\) and \(a\), the distance of the centre of mass of the letter P ,
   from (i) \(O E\) (ii) \(O A\) The letter P is freely suspended from \(O\) and hangs in equilibrium. The angle between \(O E\) and the downward vertical is \(\alpha\). Using the model,
2. find the exact value of \(\tan \alpha\)

5. \begin{figure}[h] \end{figure} The uniform plane lamina shown in Figure 3 is formed from two squares, \(A B C O\) and \(O D E F\), and a sector \(O D C\) of a circle with centre \(O\). Both squares have sides of length \(3 a\) and \(A O\) is perpendicular to \(O F\). The radius of the sector is \(3 a\) [0pt] [In part (a) you may use, without proof, any of the centre of mass formulae given in the formulae booklet.]

1. Show that the distance of the centre of mass of the sector \(O D C\) from \(O C\) is \(\frac { 4 a } { \pi }\)
2. Find the distance of the centre of mass of the lamina from \(F C\) The lamina is freely suspended from \(F\) and hangs in equilibrium with \(F C\) at an angle \(\theta ^ { \circ }\) to the downward vertical.
3. Find the value of \(\theta\)