# Lesson Worksheet: The Negative Mass Method Mathematics

In this worksheet, we will practice finding the center of gravity of a lamina that contains holes using the negative mass method.

Q1:

A uniform lamina is 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,
• B,
• C,
• D,

Q2:

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 cm
• B cm
• C cm
• D cm

Q3:

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 measure 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
• B
• C
• D

Q4:

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 .

• A11 cm
• B cm
• C22 cm
• D cm

Q5:

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 cm
• B cm
• C cm
• D cm

Q6:

An equilateral triangle 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 . When the mass at vertex is removed, the centre of mass of the system is . Find the coordinates of the centre of mass of the two systems and . • A,
• B,
• C,
• D,

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 was cut out and stuck onto square , determine the coordinates of the center of gravity of the resulting lamina. • A
• B
• C
• D

Q8:

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
• B
• C
• D

Q9:

A uniform lamina is 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. • A,
• B,
• C,
• D,

Q10:

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 masses 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. • A
• B
• C
• D

This lesson includes 15 additional questions and 213 additional question variations for subscribers.

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