Force and motion Forces acting together If several forces act on an object parallel to a given direction, then the net force is the sum of the forces in that direction minus the sum of forces in the opposite direction. 200 N 1200 N Net force = 1200 – 200 = 1000 N The net force is equal to the product of the mass of the object and its acceleration in given direction. F = ma

Force and motion Example H heavy box of mass 40 kg has a handle on one side. Two girls try to move it cross the floor. One pulls horizontally on the handle with a force of 30 N, the other pushes from the other side of the box with a force of 35 N, but the box does not move, find the frictional force resisting the motion. 35 N 30 N 40 kg R N Since the box remains at rest 30 + 35 – R = So R = 65 N

Force and motion A car of mass 1000 kg is moving with a constant speed of 25 m/s in a horizontal straight line, against a resisting force of 400 N. (a) What driving force is being provided to sustain this motion? (b) The driver speeds up uniformly over the next 20 seconds to reach a speed of 35 m/s. Assuming that the resisting force remains at 400 N, calculate the new driving force. Driving force 400 N (a) No acceleration  Driving force = 400 N (b) Data u = 25 v = 35 t = 20 a? Required equation: v = u + at 35 = 25 + 20a  a = 0.5 ms-1 D – 400 = 1000 0.5  D = 900 N Driving force = 900 N

Force and motion Example Two builders push a skip of mass 400 kg across the ground. They both push horizontally, one with a force of 300 N, the other with 350 N. Motion is resisted by a frictional force of 490 . Find the acceleration of the skip. a ms-2 300 N 400 kg 350 N 490 N Net force = 300 + 350 – 490 = 160 N F = ma 160 = 400 a a = 0.4 ms-2

Force and motion Example A car of mass 800 kg is travelling a straight horizontal road with an acceleration of 2 ms-2. The engine is exerting a forward force of magnitude 1800. By modelling the car as a particle, find the magnitude of the resistance it is experiencing. 2 ms-2 800 kg 1800 N R N Net force = 1800 – R F = m a 1800 – R = 800 a = 800  2 R = 200 N

Force and motion Example A dinghy of mass 50 kg is moved across a horizontal beach at a steady speed of 2 ms-1. One of the crew pulls with a force P, the other pushes with a force of (P + 10) newtons. The frictional force resisting the motion is 100 N. Find P. direction of motion Net force = P + P + 10 - 100 0 ms-2 P N (P+10) N Constant velocity 50 kg P + P + 10 – 100 = 100 N 2P – 90 = 0 P = 45 N

Force and motion Example A wagon of mass 200 kg is pulled by a horizontal cable along a straight level track against a resistance force of 150 N. The wagon starts from rest. After 5 seconds it has cover a distance of 30m. Find the tension in the cable. F = ma a ms-2 T – 150 = 200 a 200 kg T N 150 N Data: u = 0 t =5 s = 30 a? Required equation: s = ut + ½ at2 30 = ½  a  52 gives: a = 2.4 T – 150 = 200  2.4 Gives T = 630 N