Force is a vector quantity Its units are newton's N Force as a Vector Force is a vector quantity Its units are newton's N

Resultant Force A resultant force is the single force which represents the vector sum of two or more forces. For example the two forces of magnitude 𝐹 1 and 𝐹 2 acting upon a particle have a resultant as shown. Resultant 𝐹 1 𝐹 2 𝐹 1 Resultant 𝐹 2

Resultant Force In order to add two force together the “start “ of the second force needs to be moved to the end of the first force with the resultant going from the start of the first force directly to the end of the second force as shown in the diagram below. 𝐹 1 𝐹 2 Resultant

Worked Example 1 Consider two forces of magnitude F1 = 5N and F2 =7N acting on a particle with an angle of 90 0 between them. What is the magnitude and direction of the resultant force? Solution 𝐹𝑟 2 = 𝐹1 2 + 𝐹2 2 𝐹𝑟 2 = 5 2 + 7 2 𝐹𝑟 2 =74 𝐹𝑟= 74 𝐹𝑟=8.6 5N 5N 7N 7N

Worked Example 1 How do we work out the angle ? What about trigonometry ? SOHCAHTOA tan −1 = 𝑂𝑝𝑝 𝐴𝑑𝑗 tan −1 = 5 7 𝑡𝑎𝑛 = 35.537 0 Hence the resultant has a magnitude of 8.6N at 35.537 0 above the positive x-axis

Student Example Consider two force of magnitude of 3N and 4N with an angle of 90 0 . What is the magnitude and direction of the resultant force. 3N 4N 𝐹𝑟 2 = 𝐹1 2 + 𝐹2 2 𝐹𝑟 2 = 3 2 + 4 2 𝐹𝑟 2 =9+14 Fr= 25 =5

tan −1 = 𝑂𝑝𝑝 𝐴𝑑𝑗 tan −1 = 3 4 𝑡𝑎𝑛 = 36.869 0 Hence the resultant has a magnitude of 5N at 36.869 0 above the positive x-axis

What if the angle is not 90 0 We need to use the Cosine and Sine rules

Consider two forces of magnitude 3N and 4N with an angle of 𝛼 𝛼= 30 0 Solution C 3N B 4N A

How do we calculate the angle We need to use the Sine Rule What do we know C 3N 𝛼= 150 0 B 4N A

Since we need to find Sin B we will use the following Transpose for Sin B

Student Questions 1. Add the following vectors and determine the resultant. 3.0 m/s, 45 deg and 5.0 m/s, 135 deg + 2. Add the following vectors and determine the resultant. 5.0 m/s, 45 deg and 2.0 m/s, 180 deg 135 0 45 0 180 0 45 0

Answers Question 1 𝐹𝑟 2 = 𝐹1 2 + 𝐹2 2 𝐹𝑟 2 = 3 2 + 5 2 𝐹𝑟 2 =9+25 The resultant is magnitude 5.83𝑚/𝑠 and direction ( 45 0 + 59.036 0 )= 105 0 𝐹𝑟 2 = 𝐹1 2 + 𝐹2 2 𝐹𝑟 2 = 3 2 + 5 2 𝐹𝑟 2 =9+25 Fr= 34 =5.83𝑚/𝑠 tan −1 = 𝑂𝑝𝑝 𝐴𝑑𝑗 tan −1 = 5 3 𝑡𝑎𝑛 = 59.036 0

Answers Question 2

The resultant is magnitude 3. 854𝑚/𝑠 and direction ( 21