Worksheet: The Negative Mass Method

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

Q1:

A uniform lamina in the form of a rectangle 𝐴𝐵𝐶𝐷 where 𝐴𝐵=37cm and 𝐴𝐷=23cm. Two points 𝐸 and 𝐹 lie on 𝐵𝐷 such that 𝐵𝐹=10cm and 𝐷𝐸=15cm. A hole of radius 5 cm is drilled at 𝐹 and another of radius 4 cm is drilled at 𝐸. Find the coordinates of the point 𝑁 on 𝐴𝐵 from which the lamina can be hung so that 𝐴𝐵 is horizontal when hanging in its equilibrium position. Secondly, find the coordinates of the point 𝐾 on 𝐴𝐷 from which the lamina can be hung so that 𝐴𝐷 is horizontal when hanging in its equilibrium position. Round your answers to two decimal places if necessary.

  • A𝑁(23,58.86), 𝐾(50.81,37)
  • B𝑁(23,36.76), 𝐾(10.32,37)
  • C𝑁(23,19.19), 𝐾(11.07,37)
  • D𝑁(23,237), 𝐾(24.62,37)

Q2:

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

  • A27310 cm
  • B27104 cm
  • C274712 cm
  • D275710 cm

Q3:

A uniform lamina is in the form of a rectangle 𝐴𝐵𝐶𝐷 in which 𝐴𝐵=64cm and 𝐵𝐶=240cm. The corner 𝐴𝐵𝐸, where 𝐸 is the midpoint of 𝐴𝐷, was cut off. The resulting lamina 𝐴𝐶𝐷𝐸 was freely suspended from the vertex 𝐶. Determine the measure of the angle the side 𝐶𝐵 makes to the vertical when the lamina is hanging in its equilibrium position stating your answer to the nearest minute.

  • A1657
  • B733
  • C8224
  • D736

Q4:

A uniform rectangular lamina 𝐴𝐵𝐶𝐷 has side lengths 𝐴𝐵=24cm and 𝐵𝐶=11cm. A straight cut was made from the point 𝐸 on side 𝐵𝐶 to the point 𝐹 on side 𝐵𝐴, dividing the lamina into the triangular lamina 𝐵𝐸𝐹 and the pentagonal lamina 𝐴𝐹𝐸𝐶𝐷. When 𝐴𝐹𝐸𝐶𝐷 is stood on edge 𝐶𝐸, it is on the point of tipping over about 𝐸. Given that 𝐵𝐸=6cm, find the distance 𝐵𝐹.

  • A11 cm
  • B112 cm
  • C22 cm
  • D227 cm

Q5:

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

  • A6056 cm
  • B8476 cm
  • C6058 cm
  • D8478 cm

Q6:

An equilateral triangle 𝐴𝐵𝐶 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 𝐺. When the mass at vertex 𝐶 is removed, the centre of mass of the system is 𝐺. Find the coordinates of the centre of mass of the two systems 𝐺 and 𝐺.

  • A𝐺41,8233, 𝐺412,8233
  • B𝐺41,8233, 𝐺412,4132
  • C𝐺41,4133, 𝐺412,4132
  • D𝐺41,4133, 𝐺412,8233

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

  • A6910,8110
  • B233,9
  • C253,9
  • D698,818

Q8:

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

  • A(46.5,35.8)
  • B(46.5,33.56)
  • C(33.56,46.5)
  • D(35.8,46.5)

Q9:

A uniform lamina is in the form of a rectangle 𝐴𝐵𝐶𝐷, where 𝐴𝐵=56cm and 𝐵𝐶=35cm. Two points 𝐸 and 𝐹 are on 𝐴𝐵 such that 𝐴𝐸=𝐵𝐹=14cm. The triangle 𝑀𝐸𝐹, where 𝑀 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 𝐷, find the tangent of the angle that 𝐷𝐴 makes to the vertical, tan𝜃, when the lamina is hanging in its equilibrium position.

  • A28,956, tan𝜃=16895
  • B956,28, tan𝜃=16895
  • C28,956, tan𝜃=95168
  • D956,28, tan𝜃=95168

Q10:

A uniform square-shaped lamina 𝐴𝐵𝐶𝐷 of side length 222 cm has a mass of one kilogram. The midpoints of 𝐴𝐷, 𝐴𝐵, and 𝐵𝐶 are denoted by 𝑇, 𝑁, and 𝐾 respectively. The corners 𝑇𝐴𝑁 and 𝑁𝐵𝐾 were folded over so that they lie apartment on the surface of the lamina. Bodies of mass 365 g and 294 g were attached to the points 𝑇 and 𝐾 respectively. Find the coordinates of the center of mass of the system rounding your answer to two decimal places if necessary.

  • A(144.61,111.83)
  • B(161.18,106.25)
  • C(105.42,134.17)
  • D(105.42,106.25)

Q11:

A uniform circular disc with center 𝑁 has a radius of 72 cm. Point 𝐹 is 36 cm from the center. A line is drawn through perpendicular to 𝐹𝑁 which touches the edge of the disc at 𝑆 and 𝑇. Two circular holes of radius 12 cm are cut from the disc such that their centers lie on 𝑆𝑇 and they meet at 𝐹. Determine the distance 𝑑 between 𝑁 and the center of mass of the resulting shape. The disc was then freely suspended from 𝑍, the point where the radius perpendicular to 𝐹𝑁 meets the edge of the disc. When the disc is hanging in equilibrium, 𝑍𝑁 makes an angle 𝜃 with the vertical. Determine tan𝜃.

  • A𝑑=3617cm, tan𝜃=34
  • B𝑑=1217cm, tan𝜃=1102
  • C𝑑=3617cm, tan𝜃=134
  • D𝑑=1217cm, tan𝜃=102

Q12:

A square-shaped uniform lamina 𝐴𝐵𝐶𝐷 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 𝐴𝐵 and 𝐵𝐶. Determine the coordinates of the center of mass of the resulting lamina. Take 𝜋=227.

  • A22115,19915
  • B597112,458
  • C454,59756
  • D19930,22130

Q13:

A uniform rectangular lamina 𝐴𝐵𝐶𝐷 in which 𝐴𝐵=24cm and 𝐵𝐶=48cm is in the first quadrant of a Cartesian plane such that 𝐵 is at the origin and 𝐶 lies on the 𝑥-axis. The point 𝑁 lies on the edge 𝐴𝐷 such that 𝐷𝑁=32cm. The triangle 𝑁𝐶𝐷 was cut out of the lamina. Find the coordinates of the center of gravity of the system.

  • A1963,10
  • B803,14
  • C523,10
  • D523,28

Q14:

A uniform rectangular-shaped lamina 𝐴𝐵𝐶𝐷, where 𝐴𝐵=14cm and 𝐵𝐶=15cm, has a mass of 𝑚. Its center is denoted by 𝑁. The triangle 𝑁𝐴𝐵 is cut out of the lamina. Masses of 3𝑚, 5𝑚, 2𝑚, 2𝑚, and 38𝑚 are attached to the resulting lamina at points 𝐴, 𝐵, 𝐶, 𝐷, and 𝑁 respectively. Determine the coordinates of the center of mass of the system, rounding your answer to two decimal places if necessary.

  • A(9.24,6.49)
  • B(9.69,5.93)
  • C(9.69,3.84)
  • D(2.83,5.93)

Q15:

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

  • A4121,4721, tan𝜃=4147
  • B4721,4121, tan𝜃=4147
  • C4121,4721, tan𝜃=4741
  • D4721,4121, tan𝜃=4741

Q16:

A uniform triangular lamina 𝐴𝐵𝐶 is such that 𝑚𝐵=90, 𝐴𝐵=20cm, and 𝐵𝐶=27cm. A circular hole of radius 3 cm centred at the point of intersection of the medians of triangle 𝐴𝐵𝐶 was cut out of the lamina. Point 𝐷 is on the line 𝐴𝐵 such that 𝐴𝐷=5cm. Another cut was made starting from point 𝐷, cutting parallel to the base line 𝐵𝐶, to meet 𝐴𝐶 at 𝐸. The triangle 𝐴𝐷𝐸 was removed. Find the coordinates of the centre of mass of the resulting lamina rounding your answer to two decimal places, if necessary.

  • A(9.86,14.83)
  • B(9.51,5.92)
  • C(14.83,9.86)
  • D(5.92,9.51)

Q17:

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

  • A178,178
  • B1716,1716
  • C8516,8516
  • D1716,1716

Q18:

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

  • A(9,8)
  • B18922,8411
  • C27922,4
  • D92,12411
  • E92,4

Q19:

A uniform lamina in the form of a rectangle 𝐴𝐵𝐶𝐷 where 𝐴𝐵=41cm and 𝐴𝐷=47cm. Two points 𝐸 and 𝐹 lie on 𝐵𝐷 such that 𝐵𝐹=15cm and 𝐷𝐸=23cm. A hole of radius 8 cm is drilled at 𝐹 and another of radius 6 cm is drilled at 𝐸. Find the coordinates of the point 𝑁 on 𝐴𝐵 from which the lamina can be hung so that 𝐴𝐵 is horizontal when hanging in its equilibrium position. Secondly, find the coordinates of the point 𝐾 on 𝐴𝐷 from which the lamina can be hung so that 𝐴𝐷 is horizontal when hanging in its equilibrium position. Round your answers to two decimal places if necessary.

  • A𝑁(47,49.54), 𝐾(104.58,41)
  • B𝑁(47,41.48), 𝐾(20.57,41)
  • C𝑁(47,21.45), 𝐾(22.41,41)
  • D𝑁(47,365.02), 𝐾(50.49,41)

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