1 / 7

Forces acting together

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.

lloyd
Download Presentation

Forces acting together

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 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

  2. 35 N 30 N 40 kg R N 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. Since the box remains at rest 30 + 35 – R = 0 So R = 65 N

  3. 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

  4. a ms-2 300 N 350 N 400 kg 490 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. Net force = 300 + 350 – 490 = 160 N F = ma 160 = 400a a = 0.4 ms-2

  5. 2 ms-2 1800 N R N 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. 800 kg Net force = 1800 – R F = m a 1800 – R = 800 a = 800  2 R = 200 N

  6. direction of motion 0 ms-2 P N (P+10) N 50 kg 100 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. Net force = P + P + 10 - 100 Constant velocity P + P + 10 – 100 = 0 2P – 90 = 0 P = 45 N

  7. Force and motion a ms-2 200 kg T N 150 N 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 T – 150 = 200 a 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

More Related