Worksheet: The Negative Mass Method

A uniform lamina is in the form of a rectangle in which and . The corner , where is the midpoint of , was cut off. The resulting lamina was freely suspended from the vertex . Determine the size of the angle the side makes to the vertical when the lamina is hanging in its equilibrium position stating your answer to the nearest minute.

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

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

A square-shaped uniform lamina 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 and . Determine the coordinates of the center of mass of the resulting lamina. Take .

A uniform wire of length 135 cm was bent around five sides of a regular hexagon . Determine the distance between the center of gravity of the wire and the center of the hexagon.

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

A uniform rectangular lamina has side lengths and . A straight cut was made from the point on side to the point on side , dividing the lamina into the triangular lamina and the pentagonal lamina . When is stood on edge , it is on the point of tipping over about . Given that , find the distance .

A uniform rectangular lamina in which , lies is in the first quadrant of a Cartesian plane such that is at the origin and lies on the -axis. The point lies on the edge such that . The triangle was cut out of the lamina. Find the coordinates of the center of gravity of the system.

A uniform lamina in the form of a rectangle where and . Two points and are on such that . The triangle , where is the center of the rectangle, is cut out of the lamina. Find the coordinates of the center of mass of the resulting lamina. Given that the lamina was freely suspended from , find the tangent of the angle that makes to the vertical, , when the lamina is hanging in its equilibrium position.

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

A uniform circular disc with center has a radius of 72 cm. Point is 36 cm from the center. A line is draw perpendicular to which touches the edge of the disc at and . Two circular holes of radius 12 cm are drilled into the disc such that their centers lie on and they meet at . Determine the distance between and the center of gravity of the resulting shape. The disc was then freely suspended from a point which is where the perpendicular radius to meets the edge of the disc. Determine the tangent of the angle that makes to the vertical, , when the disc is hanging in its equilibrium position.

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

A uniform lamina in the form of a rectangle where and . Two points and lie on such that and . A hole of radius 5 cm is drilled at and another of radius 4 cm is drilled at . Find the coordinates of the point on from which the lamina can be hung so that is horizontal when hanging in its equilibrium position. Secondly, find the coordinates of the point on from which the lamina can be hung so that is horizontal when hanging in its equilibrium position. Round your answers to two decimal places if necessary.

A uniform lamina in the form of a rectangle where and . Two points and lie on such that and . A hole of radius 8 cm is drilled at and another of radius 6 cm is drilled at . Find the coordinates of the point on from which the lamina can be hung so that is horizontal when hanging in its equilibrium position. Secondly, find the coordinates of the point on from which the lamina can be hung so that is horizontal when hanging in its equilibrium position. Round your answers to two decimal places if necessary.

A uniform square-shaped lamina of side length 222 cm has a mass of one kilogram. The midpoints of , , and are denoted by , , and respectively. The corners and 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 and respectively. Find the coordinates of the center of mass of the system rounding your answer to two decimal places if necessary.

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 was cut out and stuck onto square , determine the coordinates of the centre of gravity of the resulting lamina.

A uniform rectangular-shaped lamina where and has a mass of . Its center is denoted by . The triangle is cut out of the lamina. Masses of , , , , and are attached to the resulting lamina at points , , , , and respectively. Determine the coordinates of center of mass of the system rounding your answer to two decimal places, if necessary.

A uniform triangular lamina is such that , , and . A circular hole of radius 3 cm centered at the point of intersection of the medians of triangle was cut out of the lamina. Point is on the line such that . Another cut was made starting from point , cutting parallel to the base line , to meet at . The triangle was removed. Find the coordinates of the center of mass of the resulting lamina rounding your answer to two decimal places, if necessary.

An equilateral triangle has a side length of 82 cm. When three equal masses are placed at the vertices of the triangle, the center of mass of the system is . When the mass at vertex is removed, the center of mass of the system is . Find the coordinates of the center of mass of the two systems and .