Lesson Worksheet: The Negative Mass Method Mathematics

Start Practising

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 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.

  • A27√310 cm
  • B27√104 cm
  • C27√4712 cm
  • D27√5710 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 𝐵𝐶𝐷𝐸 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′.

  • A𝐺41,82√33, 𝐺′412,82√33
  • B𝐺41,82√33, 𝐺′412,41√32
  • C𝐺41,41√33, 𝐺′412,41√32
  • D𝐺41,41√33, 𝐺′412,82√33

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 flat on the surface of the lamina. Bodies of masses 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(105.42,115.75)

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

GET YOUR PORTAL NOW

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.