Equilibrium with applied force

3 \(A B C D E F\) is the L -shaped cross-section of a uniform solid. This cross-section passes through the centre of mass of the solid and has dimensions as shown in Fig. 1.

1. Find the distance of the centre of mass of the solid from the edge \(A B\) of the cross-section.
   [diagram]
   The solid rests in equilibrium with the face containing the edge \(A F\) of the cross-section in contact with a horizontal table. The weight of the solid is \(W\) N. A horizontal force of magnitude \(P\) N is applied to the solid at the point \(B\), in the direction of \(B C\) (see Fig. 2). The table is sufficiently rough to prevent sliding.
2. Find \(P\) in terms of \(W\), given that the equilibrium of the solid is about to be broken.

2 \includegraphics[max width=\textwidth, alt={}, center]{5a2248f6-3ef9-4e69-90cf-4d6a2351be14-2_319_908_438_616} A uniform solid cone has height 20 cm and base radius \(4 \mathrm {~cm} . P Q\) is a diameter of the base of the cone. The cone is held in equilibrium, with \(P\) in contact with a horizontal surface and \(P Q\) vertical, by a force applied at \(Q\). This force has magnitude 3 N and acts parallel to the axis of the cone (see diagram). Calculate the mass of the cone.

6 \includegraphics[max width=\textwidth, alt={}, center]{6d3892e0-8c88-44ec-940f-c526d71a7fc6-3_720_723_1165_712} The diagram shows the cross-section \(O A B C D E\) through the centre of mass of a uniform prism. The interior angles of the cross-section at \(O , A , C , D\) and \(E\) are all right angles. \(O A = 0.4 \mathrm {~m} , A B = 0.5 \mathrm {~m}\) and \(B C = C D = 1 \mathrm {~m}\).

1. Calculate the distance of the centre of mass of the prism from \(O E\). The weight of the prism is 120 N . A force of magnitude \(F \mathrm {~N}\) acting along \(D E\) holds the prism in equilibrium when \(O A\) rests on a rough horizontal surface.
2. Find the set of possible values of \(F\).

4 \includegraphics[max width=\textwidth, alt={}, center]{f8633b64-b20c-4471-9641-ccc3e6854f2c-3_784_556_260_790} A uniform object is made by drilling a cylindrical hole through a rectangular block. The axis of the cylindrical hole is perpendicular to the cross-section \(A B C D\) through the centre of mass of the object. \(A B = C D = 0.7 \mathrm {~m} , B C = A D = 0.4 \mathrm {~m}\), and the centre of the hole is 0.1 m from \(A B\) and 0.2 m from \(A D\) (see diagram). The hole has a cross-section of area \(0.03 \mathrm {~m} ^ { 2 }\).

1. Show that the distance of the centre of mass of the object from \(A B\) is 0.212 m , and calculate the distance of the centre of mass from \(A D\). The object has weight 70 N and is placed on a rough horizontal surface, with \(A D\) in contact with the surface. A vertically upwards force of magnitude \(F \mathrm {~N}\) acts on the object at \(C\). The object is on the point of toppling.
2. Find the value of \(F\). The force acting at \(C\) is removed, and the object is placed on a rough plane inclined at an angle \(\theta ^ { \circ }\) to the horizontal. \(A D\) lies along a line of greatest slope, with \(A\) higher than \(D\). The plane is sufficiently rough to prevent sliding, and the object does not topple.
3. Find the greatest possible value of \(\theta\).

4 \includegraphics[max width=\textwidth, alt={}, center]{0a80f46b-b37e-46ce-8907-9d10e4f62f6d-3_388_650_264_749} The diagram shows the cross-section \(A B C D\) through the centre of mass of a uniform solid prism. \(A B = 0.9 \mathrm {~m} , B C = 2 a \mathrm {~m} , A D = a \mathrm {~m}\) and angle \(A B C =\) angle \(B A D = 90 ^ { \circ }\).

1. Calculate the distance of the centre of mass of the prism from \(A D\).
2. Express the distance of the centre of mass of the prism from \(A B\) in terms of \(a\). The prism has weight 18 N and rests in equilibrium on a rough horizontal surface, with \(A D\) in contact with the surface. A horizontal force of magnitude 6 N is applied to the prism. This force acts through the centre of mass in the direction \(B C\).
3. Given that the prism is on the point of toppling, calculate \(a\).

4. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E\) is made by joining a rectangular lamina \(A B D E\) to a triangular lamina \(B C D\) along the edge \(B D\). The rectangle has length \(6 a\) and width \(3 a\). The triangle is isosceles, with \(B C = C D\), and the distance from \(C\) to \(B D\) is \(3 a\), as shown in Figure 2.

1. Find the distance of the centre of mass of the lamina, \(A B C D E\), from \(A E\). The lamina \(A B C D E\) is freely suspended from \(A\). A horizontal force of magnitude \(F\) newtons is applied to the lamina at \(D\). The line of action of the force lies in the vertical plane containing the lamina. The lamina is in equilibrium with \(A E\) vertical. The mass of the lamina is 4 kg .
2. Find the magnitude of the force exerted on the lamina at \(A\).
   DO NOT WIRITE IN THIS AREA

3 \includegraphics[max width=\textwidth, alt={}, center]{dd23f4a8-f7e7-4f80-bad6-7e9ec21565fc-2_828_476_1338_836} A door is modelled as a lamina \(A B C D E\) consisting of a uniform rectangular section \(A B D E\) of weight 60 N and a uniform semicircular section \(B C D\) of weight 10 N and radius \(40 \mathrm {~cm} . A B\) is 200 cm and \(A E\) is 80 cm . The door is freely hinged at \(F\) and \(G\), where \(G\) is 30 cm below \(B\) and \(F\) is 30 cm above \(A\) (see diagram).

1. Find the magnitudes and directions of the horizontal components of the forces on the door at each of \(F\) and \(G\).
2. Calculate the distance from \(A E\) to the centre of mass of the door.

3 A uniform rectangular lamina \(A B C D\), of mass 1.6 kg , has side \(A B\) of length 12 cm and side \(B C\) of length 8 cm . To create a logo, a uniform circular lamina, of mass 0.4 kg , is attached. The centre of the circular lamina is at the point \(C\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{088327c1-acd3-486d-b76f-1fe2560ffaff-3_520_780_593_630}

1. Find the distance of the centre of mass of the logo:
   1. from the line \(A B\);
   2. from the line \(A D\).
2. The logo is suspended in equilibrium, with \(A B\) horizontal, by two vertical strings. One string is attached at the point \(A\) and the other string is attached at the point \(B\). Find the tension in each of the two strings.

8

1. Fig. 1 A uniform lamina \(A B C D\) is in the form of a right-angled trapezium. \(A B = 6 \mathrm {~cm} , B C = 8 \mathrm {~cm}\) and \(A D = 17 \mathrm {~cm}\) (see Fig. 1). Taking \(x\) - and \(y\)-axes along \(A D\) and \(A B\) respectively, find the coordinates of the centre of mass of the lamina.
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6ae57fe9-3b6f-46c2-95b8-d48903ed796b-5_481_1079_991_575} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} The lamina is smoothly pivoted at \(A\) and it rests in a vertical plane in equilibrium against a fixed smooth block of height 7 cm . The mass of the lamina is 3 kg . \(A D\) makes an angle of \(30 ^ { \circ }\) with the horizontal (see Fig. 2). Calculate the magnitude of the force which the block exerts on the lamina.

4 \includegraphics[max width=\textwidth, alt={}, center]{8492ec9b-3327-4d89-aaa4-bf98cdf0ebdc-2_721_513_1260_762} A uniform solid prism has cross-section \(A B C D E\) in the shape of a rectangle measuring 20 cm by 4 cm joined to a semicircle of radius 8 cm as shown in the diagram. The centre of mass of the solid lies in this cross-section.

1. Find the distance of the centre of mass of the solid from \(A B\). The solid is placed with \(A E\) on rough horizontal ground (so the object does not slide) and is in equilibrium with a horizontal force of magnitude 4 N applied along \(C B\).
2. Find the greatest and least possible values for the weight of the solid.

3 \includegraphics[max width=\textwidth, alt={}, center]{b96a99a6-3df4-4000-9bf1-aab7ab954b4a-2_389_698_1706_694} A uniform lamina \(A B C D\) has the shape of a square of side \(a\) adjoining a right-angled isosceles triangle whose equal sides are also of length \(a\). The weight of the lamina is \(W\). The lamina rests, in a vertical plane, on smooth supports at \(A\) and \(D\), with \(A D\) horizontal (see diagram).

1. Show that the centre of mass of the lamina is at a horizontal distance of \(\frac { 11 } { 9 } a\) from \(A\).
2. Find, in terms of \(W\), the magnitudes of the forces on the supports at \(A\) and \(D\).

5 A uniform lamina OABC is in the shape of a trapezium where O is the origin of the coordinate system in which the points \(\mathrm { A } , \mathrm { B }\) and C have coordinates \(( 120,0 )\), \(( 60,90 )\) and \(( 30,90 )\) respectively (see diagram). The units of the axes are centimetres. \includegraphics[max width=\textwidth, alt={}, center]{0a790ad0-7eda-40f1-9894-f156766ae46f-5_561_720_404_242} The centre of mass of the lamina lies at ( \(\mathrm { x } , \mathrm { y }\) ).

1. Show that \(\bar { x } = 54\) and determine the value of \(\bar { y }\). The lamina is placed horizontally so that it rests on three supports, whose points of contact are at \(\mathrm { B } , \mathrm { C }\) and D , where D lies on the edge OA and has coordinates \(( d , 0 )\).
2. Determine the range of values of \(d\) for the lamina to rest in equilibrium. It is now given that \(d = 63\), and that the lamina has a weight of 100 N .
3. Determine the forces exerted on the lamina by each of the supports at \(\mathrm { B } , \mathrm { C }\) and D .

3. \begin{figure}[h] \end{figure} The uniform lamina \(A B C D E F G H I J\) is shown in Figure 3.
The lamina has \(A J = 8 a , A B = 5 a\) and \(B C = D E = E F = F G = G H = H I = I J = 2 a\).
All the corners are right angles.

1. Show that the distance of the centre of mass of the lamina from \(A J\) is \(\frac { 49 } { 26 } a\) A light inextensible rope is attached to the lamina at \(A\) and another light inextensible rope is attached to the lamina at \(B\). The lamina hangs in equilibrium with both ropes vertical and \(A B\) horizontal. The weight of the lamina is \(W\).
2. Find, in terms of \(W\), the tension in the rope attached to the lamina at \(B\). The rope attached to \(B\) breaks and subsequently the lamina hangs freely in equilibrium, suspended from \(A\).
3. Find the size of the angle between \(A J\) and the downward vertical.

1. \begin{figure}[h] \end{figure} A uniform plane lamina is in the shape of an isosceles trapezium \(A B C D E F\), as shown shaded in Figure 1.

• \(B C E F\) is a square
• \(A B = C D = a\)
• \(B C = 3 a\)
  1. Show that the distance of the centre of mass of the lamina from \(A D\) is \(\frac { 11 a } { 8 }\)

The mass of the lamina is \(M\) The lamina is suspended by two light vertical strings, one attached to the lamina at \(A\) and the other attached to the lamina at \(F\) The lamina hangs freely in equilibrium, with \(B F\) horizontal.

Find, in terms of \(M\) and \(g\), the tension in the string attached at \(A\)

4. \begin{figure}[h] \end{figure} A uniform triangular lamina \(A B C\) is isosceles, with \(A C = B C\). The midpoint of \(A B\) is \(M\). The length of \(A B\) is \(18 a\) and the length of \(C M\) is \(18 a\). The triangular lamina \(C D E\), with \(D E = 6 a\) and \(C D = 12 a\), has \(E D\) parallel to \(A B\) and \(M D C\) is a straight line. Triangle \(C D E\) is removed from triangle \(A B C\) to form the lamina \(L\), shown shaded in Figure 1. The distance of the centre of mass of \(L\) from \(M C\) is \(d\).

1. Show that \(d = \frac { 4 } { 7 } a\) The lamina \(L\) is suspended by two light inextensible strings. One string is attached to \(L\) at \(A\) and the other string is attached to \(L\) at \(B\).
   The lamina hangs in equilibrium in a vertical plane with the strings vertical and \(A B\) horizontal.
   The weight of \(L\) is \(W\)
2. Find, in terms of \(W\), the tension in the string attached to \(L\) at \(B\) The string attached to \(L\) at \(B\) breaks, so that \(L\) is now suspended from \(A\). When \(L\) is hanging in equilibrium in a vertical plane, the angle between \(A B\) and the downward vertical through \(A\) is \(\theta ^ { \circ }\)
3. Find the value of \(\theta\)

5. \begin{figure}[h] \end{figure} A uniform lamina \(O A B\) is modelled by the finite region bounded by the \(x\)-axis, the \(y\)-axis and the curve with equation \(y = 9 - x ^ { 2 }\), for \(x \geqslant 0\), as shown shaded in Figure 3. The unit of length on both axes is 1 m . The area of the lamina is \(18 \mathrm {~m} ^ { 2 }\)

1. Show that the centre of mass of the lamina is 3.6 m from \(\boldsymbol { O B }\).
   [0pt] [ Solutions relying on calculator technology are not acceptable.] A light string has one end attached to the lamina at \(O\) and the other end attached to the ceiling. A second light string has one end attached to the lamina at \(A\) and the other end attached to the ceiling.
   The lamina hangs in equilibrium with the strings vertical and \(O A\) horizontal.
   The weight of the lamina is \(W\) The tension in the string attached to the lamina at \(A\) is \(\lambda W\)
2. Find the value of \(\lambda\)

5. \begin{figure}[h] \end{figure} A shop sign is modelled as a uniform rectangular lamina \(A B C D\) with a semicircular lamina removed. The semicircle has radius \(a , B C = 4 a\) and \(C D = 2 a\).
The centre of the semicircle is at the point \(E\) on \(A D\) such that \(A E = d\), as shown in Figure 3.

1. Show that the centre of mass of the sign is \(\frac { 44 a } { 3 ( 16 - \pi ) }\) from \(A D\). The sign is suspended using vertical ropes attached to the sign at \(A\) and at \(B\) and hangs in equilibrium with \(A B\) horizontal. The weight of the sign is \(W\) and the ropes are modelled as light inextensible strings.
2. Find, in terms of \(W\) and \(\pi\), the tension in the rope attached at \(B\). The rope attached at \(B\) breaks and the sign hangs freely in equilibrium suspended from \(A\), with \(A D\) at an angle \(\alpha\) to the downward vertical. Given that \(\tan \alpha = \frac { 11 } { 18 }\)
3. find \(d\) in terms of \(a\) and \(\pi\).