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 𝐴𝐡𝐢𝐷 where 𝐴𝐡=37cm and 𝐴𝐷=23cm. Two points 𝐸 and 𝐹 lie on 𝐡𝐷 such that 𝐡𝐹=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 𝐴𝐡 from which the lamina can be hung so that 𝐴𝐡 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.

  • A 𝑁 ( 2 3 , 5 8 . 8 6 ) , 𝐾 ( 5 0 . 8 1 , 3 7 )
  • B 𝑁 ( 2 3 , 3 6 . 7 6 ) , 𝐾 ( 1 0 . 3 2 , 3 7 )
  • C 𝑁 ( 2 3 , 1 9 . 1 9 ) , 𝐾 ( 1 1 . 0 7 , 3 7 )
  • D 𝑁 ( 2 3 , 2 3 7 ) , 𝐾 ( 2 4 . 6 2 , 3 7 )

Q2:

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

  • A 2 7 √ 3 1 0 cm
  • B 2 7 √ 1 0 4 cm
  • C 2 7 √ 4 7 1 2 cm
  • D 2 7 √ 5 7 1 0 cm

Q3:

A uniform lamina is in the form of a rectangle 𝐴𝐡𝐢𝐷 in which 𝐴𝐡=64cm and 𝐡𝐢=240cm. The corner 𝐴𝐡𝐸, where 𝐸 is the midpoint of 𝐴𝐷, was cut off. The resulting lamina 𝐴𝐢𝐷𝐸 was freely suspended from the vertex 𝐢. Determine the measure of the angle the side 𝐢𝐡 makes to the vertical when the lamina is hanging in its equilibrium position stating your answer to the nearest minute.

  • A 1 6 5 7 β€² ∘
  • B 7 3 3 β€² ∘
  • C 8 2 2 4 β€² ∘
  • D 7 3 6 β€² ∘

Q4:

A uniform rectangular lamina 𝐴𝐡𝐢𝐷 has side lengths 𝐴𝐡=24cm and 𝐡𝐢=11cm. A straight cut was made from the point 𝐸 on side 𝐡𝐢 to the point 𝐹 on side 𝐡𝐴, dividing the lamina into the triangular lamina 𝐡𝐸𝐹 and the pentagonal lamina 𝐴𝐹𝐸𝐢𝐷. When 𝐴𝐹𝐸𝐢𝐷 is stood on edge 𝐢𝐸, it is on the point of tipping over about 𝐸. Given that 𝐡𝐸=6cm, find the distance 𝐡𝐹.

  • A11 cm
  • B 1 1 2 cm
  • C22 cm
  • D 2 2 7 cm

Q5:

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

  • A 6 0 5 6 cm
  • B 8 4 7 6 cm
  • C 6 0 5 8 cm
  • D 8 4 7 8 cm

Q6:

An equilateral triangle 𝐴𝐡𝐢 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 𝐢 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′.

  • A 𝐺 ο€Ώ 4 1 , 8 2 √ 3 3  , 𝐺 β€² ο€Ώ 4 1 2 , 8 2 √ 3 3 
  • B 𝐺 ο€Ώ 4 1 , 8 2 √ 3 3  , 𝐺 β€² ο€Ώ 4 1 2 , 4 1 √ 3 2 
  • C 𝐺 ο€Ώ 4 1 , 4 1 √ 3 3  , 𝐺 β€² ο€Ώ 4 1 2 , 4 1 √ 3 2 
  • D 𝐺 ο€Ώ 4 1 , 4 1 √ 3 3  , 𝐺 β€² ο€Ώ 4 1 2 , 8 2 √ 3 3 

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

  • A ο€Ό 2 3 3 , 9 
  • B ο€Ό 6 9 1 0 , 8 1 1 0 
  • C ο€Ό 2 5 3 , 9 
  • D ο€Ό 6 9 8 , 8 1 8 

Q8:

The diagram shows a uniform lamina 𝐴𝐡𝐢 from which a triangle 𝐺𝐡𝐢 has been cut out. 𝐴𝐡𝐢 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 ( 3 5 . 8 , 4 6 . 5 )
  • B ( 4 6 . 5 , 3 5 . 8 )
  • C ( 3 3 . 5 6 , 4 6 . 5 )
  • D ( 4 6 . 5 , 3 3 . 5 6 )

Q9:

A uniform lamina is in the form of a rectangle 𝐴𝐡𝐢𝐷, where 𝐴𝐡=56cm and 𝐡𝐢=35cm. Two points 𝐸 and 𝐹 are on 𝐴𝐡 such that 𝐴𝐸=𝐡𝐹=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, tanπœƒ, when the lamina is hanging in its equilibrium position.

  • A ο€Ό 2 8 , 9 5 6  , t a n πœƒ = 1 6 8 9 5
  • B ο€Ό 9 5 6 , 2 8  , t a n πœƒ = 1 6 8 9 5
  • C ο€Ό 2 8 , 9 5 6  , t a n πœƒ = 9 5 1 6 8
  • D ο€Ό 9 5 6 , 2 8  , t a n πœƒ = 9 5 1 6 8

Q10:

A uniform square-shaped lamina 𝐴𝐡𝐢𝐷 of side length 222 cm has a mass of one kilogram. The midpoints of 𝐴𝐷, 𝐴𝐡, and 𝐡𝐢 are denoted by 𝑇, 𝑁, and 𝐾 respectively. The corners 𝑇𝐴𝑁 and 𝑁𝐡𝐾 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 ( 1 6 1 . 1 8 , 1 0 6 . 2 5 )
  • B ( 1 4 4 . 6 1 , 1 1 1 . 8 3 )
  • C ( 1 0 5 . 4 2 , 1 3 4 . 1 7 )
  • D ( 1 0 5 . 4 2 , 1 0 6 . 2 5 )

Q11:

A uniform circular disc with centre 𝑁 has a radius of 72 cm. Point 𝐹 is 36 cm from the centre. 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 centres lie on 𝑆𝑇 and they meet at 𝐹. Determine the distance 𝑑 between 𝑁 and the centre 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 πœƒ with the vertical; determine tanπœƒ.

  • A 𝑑 = 1 2 1 7 c m , t a n πœƒ = 1 1 0 2
  • B 𝑑 = 3 6 1 7 c m , t a n πœƒ = 1 3 4
  • C 𝑑 = 3 6 1 7 c m , t a n πœƒ = 3 4
  • D 𝑑 = 1 2 1 7 c m , t a n πœƒ = 1 0 2

Q12:

A square-shaped uniform lamina 𝐴𝐡𝐢𝐷 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 𝐴𝐡 and 𝐡𝐢. Determine the coordinates of the center of mass of the resulting lamina. Take πœ‹=227.

  • A ο€Ό 2 2 1 1 5 , 1 9 9 1 5 
  • B ο€Ό 5 9 7 1 1 2 , 4 5 8 
  • C ο€Ό 4 5 4 , 5 9 7 5 6 
  • D ο€Ό 1 9 9 3 0 , 2 2 1 3 0 

Q13:

A uniform rectangular lamina 𝐴𝐡𝐢𝐷 in which 𝐴𝐡=24cm and 𝐡𝐢=48cm is in the first quadrant of a Cartesian plane such that 𝐡 is at the origin and 𝐢 lies on the π‘₯-axis. The point 𝑁 lies on the edge 𝐴𝐷 such that 𝐷𝑁=32cm. The triangle 𝑁𝐢𝐷 was cut out of the lamina. Find the coordinates of the center of gravity of the system.

  • A ο€Ό 1 9 6 3 , 1 0 
  • B ο€Ό 8 0 3 , 1 4 
  • C ο€Ό 5 2 3 , 1 0 
  • D ο€Ό 5 2 3 , 2 8 

Q14:

A uniform rectangular-shaped lamina 𝐴𝐡𝐢𝐷, where 𝐴𝐡=14cm and 𝐡𝐢=15cm, has a mass of π‘š. Its center is denoted by 𝑁. The triangle 𝑁𝐴𝐡 is cut out of the lamina. Masses of 3π‘š, 5π‘š, 2π‘š, 2π‘š, and 38π‘š are attached to the resulting lamina at points 𝐴, 𝐡, 𝐢, 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 . 2 4 , 6 . 4 9 )
  • B ( 9 . 6 9 , 5 . 9 3 )
  • C ( 9 . 6 9 , 3 . 8 4 )
  • D ( 2 . 8 3 , 5 . 9 3 )

Q15:

A uniform square lamina 𝐴𝐡𝐢𝐷 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 𝐴𝐡 makes to the vertical when the lamina is hanging in its equilibrium position is denoted by πœƒ, find tanπœƒ.

  • A ο€Ό 4 1 2 1 , 4 7 2 1  , t a n πœƒ = 4 1 4 7
  • B ο€Ό 4 7 2 1 , 4 1 2 1  , t a n πœƒ = 4 1 4 7
  • C ο€Ό 4 1 2 1 , 4 7 2 1  , t a n πœƒ = 4 7 4 1
  • D ο€Ό 4 7 2 1 , 4 1 2 1  , t a n πœƒ = 4 7 4 1

Q16:

A uniform triangular lamina 𝐴𝐡𝐢 is such that ∠𝐡=90∘, 𝐴𝐡=20cm, and 𝐡𝐢=27cm. A circular hole of radius 3 cm centred at the point of intersection of the medians of triangle 𝐴𝐡𝐢 was cut out of the lamina. Point 𝐷 is on the line 𝐴𝐡 such that 𝐴𝐷=5cm. Another cut was made starting from point 𝐷, cutting parallel to the base line 𝐡𝐢, to meet 𝐴𝐢 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 . 8 6 , 1 4 . 8 3 )
  • B ( 9 . 5 1 , 5 . 9 2 )
  • C ( 1 4 . 8 3 , 9 . 8 6 )
  • D ( 5 . 9 2 , 9 . 5 1 )

Q17:

A uniform lamina 𝐴𝐡𝐢𝐷 is in the form of a square of side length 51 cm. The points 𝐸 and 𝐹 are the midpoints of 𝐴𝐡 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.

  • A ο€Ό βˆ’ 1 7 8 , βˆ’ 1 7 8 
  • B ο€Ό βˆ’ 1 7 1 6 , βˆ’ 1 7 1 6 
  • C ο€Ό 8 5 1 6 , 8 5 1 6 
  • D ο€Ό 1 7 1 6 , 1 7 1 6 

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 )
  • B ο€Ό 1 8 9 2 2 , 8 4 1 1 
  • C ο€Ό 2 7 9 2 2 , 4 
  • D ο€Ό 9 2 , 1 2 4 1 1 
  • E ο€Ό 9 2 , 4 

Q19:

A uniform lamina in the form of a rectangle 𝐴𝐡𝐢𝐷 where 𝐴𝐡=41cm and 𝐴𝐷=47cm. Two points 𝐸 and 𝐹 lie on 𝐡𝐷 such that 𝐡𝐹=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 𝐴𝐡 from which the lamina can be hung so that 𝐴𝐡 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.

  • A 𝑁 ( 4 7 , 4 9 . 5 4 ) , 𝐾 ( 1 0 4 . 5 8 , 4 1 )
  • B 𝑁 ( 4 7 , 4 1 . 4 8 ) , 𝐾 ( 2 0 . 5 7 , 4 1 )
  • C 𝑁 ( 4 7 , 2 1 . 4 5 ) , 𝐾 ( 2 2 . 4 1 , 4 1 )
  • D 𝑁 ( 4 7 , 3 6 5 . 0 2 ) , 𝐾 ( 5 0 . 4 9 , 4 1 )

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