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Learning Objective : 1. To look at the how F=ma presents itself in questions..... • • • • • Towing things Rockets Lifts Pulleys Sliding down a slope Book Reference : Pages 135-137 T Car 1000kg Caravan 800kg 8000N from the Engine 3000N Resistance 8000N from the Engine T Car 1000kg Caravan 800kg 3000N Resistance 8000N from the Engine T Car 1000kg Caravan 800kg 3000N Resistance For a rocket of mass m with thrust T at launch Weight of the Rocket mg Thrust T from the rocket motors At launch when the rocket is accelerating the unbalanced force is T – mg and hence we can apply this to “f=ma”... T - mg = ma Tension T in the cable Like before the unbalanced resultant force is applied to “f=ma” T – mg = ma But there are a number of different scenarios based upon the motion of the lift (assuming upwards is positive) 1. The lift is moving up and accelerating (a>0) T = mg + ma (T > mg) Weight of the lift mg 2. The lift is moving up at constant velocity (a=0) T=mg Tension T in the cable 3. The lift is moving up and decelerating (a<0) T = mg + ma (T < mg) 4. The lift is accelerating downwards (a<0) T = mg + ma (T < mg) 5. The lift is moving down but decelerating (a>0) T = mg + ma (T > mg) The Cable tension is more than the lift weight if: •The lift is moving up and accelerating (v>0, a>0) •The lift is moving down and decelerating (v<0, a>0) Weight of the lift mg The Cable tension is less than the lift weight if: •The lift is moving up and decelerating (v>0, a<0) •The lift is moving down and accelerating (v<0, a<0) Looking at the resultant forces on both Masses: Tension T in the cable Tension T in the cable m T – mg = ma mg Acceleration of m For m: (1) M Mg For M: Acceleration of M Mg - T = Ma (2) Since we have two equations we can remove T by substitution: (from 1 T = ma + mg, substitute into 2) Mg – (ma + mg) = Ma (M-m)g = (M +m)a ° This time it is no longer a static system in Equilibrium, the object is accelerating down the slope ° The resultant force down the slope providing the acceleration is: mg sin - F = ma The problem on the previous page can be extended to consider a vehicle with an engine which is providing an engine force FE This is simply considered when balancing all the forces which ultimately provide the acceleration FE + mg sin - F = ma We have applied “f=ma” to a number of different scenarios.... In each case it has been a matter combining the forces in the system to find the unbalanced resultant force which is causing the acceleration. Effectively this gives the “f” part of “f=ma” which can then be equated to the “ma” part The equations derived here should not be learnt, rather the this lesson should be considered a collection of worked examples