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Teach A Level Maths

Teach A Level Maths. Equilibrium. Volume 4: Mechanics 1 Equilibrium. F = 0. 8. 6. 10. Suppose we have some forces with S F = 0 . This means the sum of all the forces =0 i.e. No Resultant. If S F = 0 , the forces are said to be in equilibrium.

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Teach A Level Maths

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  1. Teach A Level Maths Equilibrium

  2. Volume 4: Mechanics 1 Equilibrium

  3. F= 0 8 6 10 Suppose we have some forces with SF= 0. This means the sum of all the forces =0 i.e. No Resultant If SF= 0, the forces are said to be in equilibrium. There is more than one way of solving problems involving forces in equilibrium.

  4. 8 q 6 10 e.g.1 Find the value of q if the 3 forces are in equilibrium. 6 8 10 Method 1: We draw the forces head-to-tail. Since there is no resultant, the 3rd force closes the triangle.

  5. 8 q 6 10 8 6 e.g.1 Find the value of q if the 3 forces are in equilibrium. 6 a a 8 10 53·1 Method 1: We draw the forces head-to-tail. Tip: The triangle of forces worked well here because 2 of the forces were perpendicular, giving a right angled triangle ( easy to solve ). We have drawn a triangle of forces. We can find either of the unknown angles in the triangle. e.g. tana= a= 53·1 ( 3 s.f. ) q = 90 + a= 143 ( 3 s.f. )

  6. 6 6 8 8 10 10 We need to be careful to notice the difference between these two triangles of forces. Forces in equilibrium Only 1 triangle shows forces in equilibrium. Which triangle is it? Ans: The 1st diagram shows the forces head-to-tail and, since they form a triangle, their sum is zero. The forces are in equilibrium.

  7. 6 6 8 8 10 10 We need to be careful to notice the difference between these two triangles of forces. Forces in equilibrium Resultant In the 2nd triangle, the 10 newton force is the resultant of the other 2 forces. It’s a good idea to use a double headed arrow.

  8. The 2nd method for solving equilibrium problems is the one we use most often. It works well with 3 or more forces.

  9. If a set of forces are in equilibrium, SX = 0 and SY = 0, i.e the sums of the perpendicular components of the forces in the X and Y direction = 0.

  10. y 4 Q 5 x 70 P e.g.2 The following forces are in equilibrium. Find the values of P and Q. If we resolve in the direction of the y-axis the only unknown will be P.

  11. y 4 Q 5 x P 4 sin70 The following forces are in equilibrium. Find the values of P and Q. Pcos70 Psin70 70 70 P Resolving: Psin70 4 - = 0 -Q -Pcos70 5 = 0 Psin70 =4 Q = 5 -Pcos70 P =  Q = 5 - 4·26cos70 P = 4·26 ( 3 s.f. )  Q = 3·54 ( 3 s.f. )

  12. sina = 0·8 cosa = 0·6 e.g.3 Given that sina = 0·8 and cosa = 0·6, show that the forces shown in the diagram are in equilibrium. y 8 x a 6 We won’t need a itself, only these trig ratios, so we will substitute without finding a. 10

  13. sina = 0·8 cosa = 0·6 e.g.3 Given that sina = 0·8 and cosa = 0·6, show that the forces shown in the diagram are in equilibrium. y 10cosa 8 10sina 10 x a a 6 10 Resolving: 10cosa X = - 10sina Y = - 6 8 = 10  0·6 - 6 =8- 10  0·8 = 0 = 0 X= 0 andY= 0, so the forces are in equilibrium.

  14. 4 7 P e.g.4 The 3 forces shown in the diagram are in equilibrium. Find the values of P and a. Decide whether you would use the triangle of forces or resolving. a The triangle of forces is the easier option.

  15. 4 7 P 4 7 e.g.4 The 3 forces shown in the diagram are in equilibrium. Find the values of P and a. P 4 a a 7 P= 8·06 ( 3 s.f. ) P2= 72+ 42 tana = a= 29·7 ( 3 s.f. ) Tip: The triangle of forces is only easy if 2 of the forces are perpendicular.

  16. A set of forces acting at a point are in equilibrium when the resultant is zero. SUMMARY We can solve problems involving forces in equilibrium in the following ways: • If there are 3 or more forces, we can form two equations using X= 0 and Y= 0 where X and Y are the sums of the components in two perpendicular directions. • If there are exactly 3 forces, we can resolve the forces or use a triangle of forces, but the triangle of forces is only easy if 2 of the forces are perpendicular.

  17. EXERCISE 1. The set of forces shown in the diagram are in equilibrium. Find the values of P and q by (a) drawing a triangle of forces, and (b) resolving the forces 6 q 3 P

  18. 6 q 3 P 3 6 EXERCISE Solution: (a) 3 P 6 q q Pythagoras’ theorem: P= 5·20 ( 3 s.f. ) P2= 62- 32 sinq= q= 30

  19. 6sinq = 3 3 6 EXERCISE Solution: (b) 6sinq 6 q 6cosq 6 3 q P Resolving: = 0 6cosq 6sinq - 3 - P = 0 P = 6cos30 sinq=  P = 5·20 ( 3 s.f. ) q= 30

  20. EXERCISE 2. The set of forces shown in the diagram are in equilibrium. Find the values of P and Q. Q P 50 4 8

  21. EXERCISE Solution: Q Q 50 50 P Qsin50 4 Qcos50 8 = 0 Resolving: Qsin50 - 8  Q = 10·4 ( 3 s.f. ) Qsin50 = 8 - 4 -Qcos50 P = 0 P = 4 +Qcos50 P= 4 + 10·4cos50  P = 10·7 ( 3 s.f. )

  22. The following page contains the summary in a form suitable for photocopying.

  23. TEACH A LEVEL MATHS – MECHANICS 1 EQUILIBRIUM Summary A set of forces acting at a point are in equilibrium when the resultant is zero. We can solve problems involving forces in equilibrium in the following ways: • If there are 3 or more forces, we can form two equations using X= 0 and Y= 0 where X and Yare the sums of the components in two perpendicular directions. • If there are exactly 3 forces, we can resolve the forces or use a triangle of forces, but the triangle of forces is only easy if 2 of the forces are perpendicular.

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