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P1 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 A mass (weight 20N) is suspended by two wires as shown in the figure: relevant distances.

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Presentation on theme: "P1 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 A mass (weight 20N) is suspended by two wires as shown in the figure: relevant distances."— Presentation transcript:

1 p1 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 A mass (weight 20N) is suspended by two wires as shown in the figure: relevant distances are also marked. T 1 and T 2 are the tensions in the wires. By resolving in the horizontal and vertical directions, write down equations involving T 1 and T 2. Express these as one matrix equation. Matrix Algebra - Tutorial 1 1.

2 p2 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 For the circuit given above, write down three equations in terms of the three currents shown, and then express these in one matrix equation. 2.

3 p3 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 A 40N weight mass is on a plane whose coefficient of friction is 0.5, and the forces affecting the mass are shown. Friction and the force F parallel to the plane keep the mass in place. R is the force applied by the mass normal to the plane, so the frictional force is 0.5R. The force up the plane is thus F + 0.5R. By resolving forces horizontally and vertically, write down equations relating F and R and then express these equations as one matrix equation. Note: cos(36.87) = 0.8 and sin(36.87) = 0.6. 3.

4 p4 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 4. A mass (weight 2N) is suspended by three wires, as shown above. Let the tensions in the wires be T 1, T 2 and T 3. The components of the tensions in the x, y and z directions are For T 1 -0.2*T 1, 0.5*T 1 and -0.1*T 1. For T 2 0, 0.7*T 2 and 0.5*T 2. For T 3 0.8*T 3, -0.7*T 3 and -0.1*T 3. Express this system as a matrix equation.

5 p5 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 5. Evaluate the following if they can be evaluated - if not explain why. a) A + Bb) A * B c) A + Ed) A - 2*B e) 5 * Cf) C * D g) A * Eh) F * C i) D T * C j) D * Ak) B T * A T l) B * A m) F T * C

6 p6 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 6. Show that, even though C is not the zero matrix, A.C = B.C and yet A does not equal B. 7.Show that A.A T is a symmetrix matrix.

7 p7 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 Answers

8 p8 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1 c) Not the same size. h) Incompatible sizes.

9 p9 RJM 16/10/02EG1C2 Engineering Maths: Matrix Algebra Tutorial 1


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