Fu = ma NOTES p.19 DYNAMICS Recall that mass (kg) acceleration (m/s2) Unbalanced Force (N) Therefore, the Newton is the unbalanced force which will cause a mass of 1 kg to have an acceleration of 1 m/s2 Free body diagrams show the direction of all forces acting on one point. REMEMBER to use the + and – sign convention!

Fu = F1 + F2 + F3 + …. At SG level we used to subtract to find the unbalanced force … 100N 40N We would say Fu = 100 – 40 = 60N At H level we should use a sign convention … 100N 40N - + So we say Fu = F1 + F2 = 100 + (-40) = +60N

- + Ff = ? Fe = 3000 N Example 1 – Friction on a horizontal plane. A car’s engine force is 3 000 N. If the mass is 900 kg and it accelerates from rest to 18 ms-1 in 14 s, what is the force of friction? 1st Diagram + sign convention - + Ff = ? 900 kg Fe = 3000 N Fu = ? +ve a = ? +ve 2nd Find “a” a = v-u t u = 0 ms-1 v = 18 ms-1 a = ? t = 14 s = 18 - 0 14 = 1.3 ms-2

3rd Calculate Fu Fu= m a Fu= ? m = 900 kg a = 1.3 ms-2 = 900 x 1.3 = 1170 N 4th Calculate Ff Fu = Fe + Ff Ff = Fu - Fe Ff = 1170 - 3000 Ff = - 1830 N So force of friction is 1830 N in opposite direction to car.

Example 2 – Lift Cable Tension. A lift cable has a tension of 9800 N when the lift is at rest. a) Determine the mass of the lift? T = 9800 N m = ? W = 9800 N (as W balances T) m = W g a) W= 9800 N m = ? g = 9.8 Nkg-1 = 9800 9.8 = 1000 kg

b) What’s the cable tension as the lift moves down at 2 ms-1? Tension is still 9800 N as constant velocity requires forces to be balanced. Determine the cable tension as the lift decelerates at 1.5 ms-2 while moving down? c) 1st a = + 1.5 ms-2 +ve m = 1000 kg T =? W = - 9800 N 2nd Fu = m a = 1000 x 1.5 Fu = ? a = 1.5 ms-2 = + 1500 N 3rd Fu = T + W T = Fu - W T = 1500 – (-9800) = + 11300 N

Problems 52 – 65 4900 N a) (i) 0.015 m/s2 (ii) 3 x 106 N b) -0.0027 m/s2 54. OA … decreasing acceleration as air resistance increases AB…constant velocity as air resistance balances the weight BC … parachute opens so dramatic deceleration to lower speed CD … constant velocity as new air resistance balances weight DE …dramatic deceleration as parachutist lands. 0.02 m/s 56. 150 N 57. a) 120 N b) -20 N a) 1200 N b) 108 m c) 2592 N a) (ii) 7.7 m/s2 b) mass decreases as fuel is burned, air resistance decreases as rocket leaves Earth’s atmosphere, weight decreases as rocket leaves Earth’s gravitational field. c) 15.9 m/s2 d) Once out of Earth’s gravitational pull weight drops to zero so engines can be switched off and Earth will continue at a constant velocity. a) 1778 kg b) 62 424 N 61. 28 600 N a) 1960 N b) 2260 N c) 1960 N d) 1660 N a) (i) 2450 N (ii) 2450 N (iii) 2950 N (iv) 1950 N b) 4.2 m/s2

Problems 52 – 65 (cont.) 63. c) An empty lift so resultant force would increase giving a higher acceleration. 51.2 N b) 37.2 N, 39.2 N, 43.2 N.

Example 3 – Towed Objects. A car tows a caravan as shown 1200 kg 1000 kg Friction against car = 200 N Friction against caravan = 500 N The car accelerates at 2 ms-2 (left!) What’s the engine force of the car? What force does the towbar exert on the caravan?

a) Free body diagram -ve +ve Total mass = 2200 kg Fe = ? Ff = -700 N a = + 2 ms-2 Fu = ? 1st Fu = ma = 2200 x (+2) = +4400 N 2nd Fu = Fe + Ff Fe = Fu - F f Fe = +4400 – (-700) = 5100 N

b) What force does the towbar exert on the caravan? is -ve is +ve Caravan only m = 1000 kg Ftowbar = ? Ff = - 500 N a = + 2 ms-2 Fu = ? 1st Fu = ma = 1000 x 2 = + 2000 N 2nd Fu = Ftowbar + F f F towbar = Fu - F f = 2000 - (-500) = + 2500 N

Problems 51 – 70.