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In this lesson, we will learn how to solve problems on the negative mass method.

Q1:

A uniform lamina in the form of a rectangle π΄ π΅ πΆ π· where π΄ π΅ = 3 7 c m and π΄ π· = 2 3 c m . Two points πΈ and πΉ lie on π΅ π· such that π΅ πΉ = 1 0 c m and π· πΈ = 1 5 c m . A hole of radius 5 cm is drilled at πΉ and another of radius 4 cm is drilled at πΈ . Find the coordinates of the point π on π΄ π΅ from which the lamina can be hung so that π΄ π΅ is horizontal when hanging in its equilibrium position. Secondly, find the coordinates of the point πΎ on π΄ π· from which the lamina can be hung so that π΄ π· is horizontal when hanging in its equilibrium position. Round your answers to two decimal places if necessary.

Q2:

A uniform wire of length 135 cm was bent around five sides of a regular hexagon π΄ π΅ πΆ π· πΈ πΉ . Determine the distance between the centre of gravity of the wire and the centre of the hexagon.

Q3:

A uniform lamina is in the form of a rectangle π΄ π΅ πΆ π· in which π΄ π΅ = 6 4 c m and π΅ πΆ = 2 4 0 c m . The corner π΄ π΅ πΈ , where πΈ is the midpoint of π΄ π· , was cut off. The resulting lamina π΄ πΆ π· πΈ was freely suspended from the vertex πΆ . Determine the measure of the angle the side πΆ π΅ makes to the vertical when the lamina is hanging in its equilibrium position stating your answer to the nearest minute.

Q4:

A uniform rectangular lamina π΄ π΅ πΆ π· has side lengths π΄ π΅ = 2 4 c m and π΅ πΆ = 1 1 c m . A straight cut was made from the point πΈ on side π΅ πΆ to the point πΉ on side π΅ π΄ , dividing the lamina into the triangular lamina π΅ πΈ πΉ and the pentagonal lamina π΄ πΉ πΈ πΆ π· . When π΄ πΉ πΈ πΆ π· is stood on edge πΆ πΈ , it is on the point of tipping over about πΈ . Given that π΅ πΈ = 6 c m , find the distance π΅ πΉ .

Q5:

A uniform square-shaped board π΄ π΅ πΆ π· has a side length of 363 cm. Its diagonals meet at π . The triangle π π΅ πΆ was cut out of the board and the remaining part was suspended from a point, πΈ , on π΄ π΅ . Given that, when the body is hanging in its equilibrium position, π΄ π΅ is horizontal, calculate the length π΄ πΈ .

Q6:

An equilateral triangle π΄ π΅ πΆ has a side length of 82 cm. When three equal masses are placed at the vertices of the triangle, the centre of mass of the system is πΊ . When the mass at vertex πΆ is removed, the centre of mass of the system is πΊ β² . Find the coordinates of the centre of mass of the two systems πΊ and πΊ β² .

Q7:

The figure shows a uniform square lamina of side length 18 cm. It is divided into nine congruent squares as shown. Given that the square πΆ was cut out and stuck onto square π΄ , determine the coordinates of the center of gravity of the resulting the lamina.

Q8:

The diagram shows a uniform lamina π΄ π΅ πΆ from which a triangle πΊ π΅ πΆ has been cut out. π΄ π΅ πΆ was an equilateral triangle with a side length of 93 cm and centre of mass πΊ . Find the coordinates of the new centre of mass of the resulting lamina. Round your answer to two decimal places if necessary.

Q9:

A uniform lamina in the form of a rectangle π΄ π΅ πΆ π· where π΄ π΅ = 5 6 c m and π΅ πΆ = 3 5 c m . Two points πΈ and πΉ are on π΄ π΅ such that π΄ πΈ = π΅ πΉ = 1 4 c m . The triangle π πΈ πΉ , where π is the centre of the rectangle, is cut out of the lamina. Find the coordinates of the centre of mass of the resulting lamina. Given that the lamina was freely suspended from π· , find the tangent of the angle that π· π΄ makes to the vertical, t a n π , when the lamina is hanging in its equilibrium position.

Q10:

A uniform square-shaped lamina π΄ π΅ πΆ π· of side length 222 cm has a mass of one kilogram. The midpoints of π΄ π· , π΄ π΅ , and π΅ πΆ are denoted by π , π , and πΎ respectively. The corners π π΄ π and π π΅ πΎ were folded over so that they lie flat on the surface of the lamina. Bodies of mass 365 g and 294 g were attached to the points π and πΎ respectively. Find the coordinates of the centre of mass of the system rounding your answer to two decimal places if necessary.

Q11:

A uniform circular disc with centre π has a radius of 72 cm. Point πΉ is 36 cm from the centre. A line is drawn through perpendicular to πΉ π which touches the edge of the disc at π and π . Two circular holes of radius 12 cm are cut from the disc such that their centres lie on π π and they meet at πΉ . Determine the distance π between π and the centre of mass of the resulting shape. The disc was then freely suspended from π , the point where the radius perpendicular to πΉ π meets the edge of the disc. When the disc is hanging in equilibrium, π π makes an angle π with the vertical; determine t a n π .

Q12:

A square-shaped uniform lamina π΄ π΅ πΆ π· has a side length of 28 cm. A circular disk of radius 7 cm was cut out of the lamina such that its center was at a distance of 17 cm from both π΄ π΅ and π΅ πΆ . Determine the coordinates of the center of mass of the resulting lamina. Take π = 2 2 7 .

Q13:

A uniform rectangular lamina π΄ π΅ πΆ π· in which π΄ π΅ = 2 4 c m , π΅ πΆ = 4 8 c m lies is in the first quadrant of a Cartesian plane such that π΅ is at the origin and πΆ lies on the π₯ -axis. The point π lies on the edge π΄ π· such that π· π = 3 2 c m . The triangle π πΆ π· was cut out of the lamina. Find the coordinates of the centre of gravity of the system.

Q14:

A uniform rectangular-shaped lamina π΄ π΅ πΆ π· where π΄ π΅ = 1 4 c m and π΅ πΆ = 1 5 c m has a mass of π . Its centre is denoted by π . The triangle π π΄ π΅ is cut out of the lamina. Masses of 3 π , 5 π , 2 π , 2 π , and 3 8 π are attached to the resulting lamina at points π΄ , π΅ , πΆ , π· , and π respectively. Determine the coordinates of centre of mass of the system rounding your answer to two decimal places, if necessary.

Q15:

A uniform square lamina π΄ π΅ πΆ π· has a side length of 4 cm. Its diagonals intersect at π . The point πΈ lies at the midpoint of π· π . The triangle πΈ π΄ π· was cut out of the lamina. Determine the coordinates of the centre of mass of the resulting lamina. The lamina was freely suspended from point π΄ . If the angle that π΄ π΅ makes to the vertical when the lamina is hanging in its equilibrium position is denoted by π , find t a n π .

Q16:

A uniform triangular lamina π΄ π΅ πΆ is such that β π΅ = 9 0 β , π΄ π΅ = 2 0 c m , and π΅ πΆ = 2 7 c m . A circular hole of radius 3 cm centred at the point of intersection of the medians of triangle π΄ π΅ πΆ was cut out of the lamina. Point π· is on the line π΄ π΅ such that π΄ π· = 5 c m . Another cut was made starting from point π· , cutting parallel to the base line π΅ πΆ , to meet π΄ πΆ at πΈ . The triangle π΄ π· πΈ was removed. Find the coordinates of the centre of mass of the resulting lamina rounding your answer to two decimal places, if necessary.

Q17:

A uniform lamina π΄ π΅ πΆ π· is in the form of a square of side length 51 cm. The points πΈ and πΉ are the midpoints of π΄ π΅ and π΄ π· , respectively. The corner π΄ πΈ πΉ has been folded over along the line πΈ πΉ so that the point π΄ aligns with π , the centre of the square, as shown in the diagram. Determine the coordinates of the centre of gravity of the lamina in this form.

Q18:

Find the coordinates of the centre of mass of the following figure, which is drawn on a grid of unit squares.

Q19:

A uniform lamina in the form of a rectangle π΄ π΅ πΆ π· where π΄ π΅ = 4 1 c m and π΄ π· = 4 7 c m . Two points πΈ and πΉ lie on π΅ π· such that π΅ πΉ = 1 5 c m and π· πΈ = 2 3 c m . A hole of radius 8 cm is drilled at πΉ and another of radius 6 cm is drilled at πΈ . Find the coordinates of the point π on π΄ π΅ from which the lamina can be hung so that π΄ π΅ is horizontal when hanging in its equilibrium position. Secondly, find the coordinates of the point πΎ on π΄ π· from which the lamina can be hung so that π΄ π· is horizontal when hanging in its equilibrium position. Round your answers to two decimal places if necessary.

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