Force in effect when car brakes A car of mass m=1200 keg is traveling at a speed of km/h. Suddenly the brakes are applied and the car is brought to a stop over a distance of mm. Assuming constant breaking force find: (1) the magnitude of the breaking force, (2) the time required to stop. (3) What will be the stopping distance if the initial speed is km/h? Solution. Most of problems from Dynamics can be seen as “two parts problem”, one involving kinematics and the other – dynamics. This is a consequence of Newton’s Second Law – Force is a product of mass and acceleration.
Acceleration by itself is a purely kinematical” problem. When mass is involved, we go into Dynamics. In our problem the following are given: m = 1200 keg – mass of the car, FL = 50 km/h – initial speed in the first case, Del = mm – stopping distance in the first case, iv = 100 km/h – initial speed in the second case. We are suppose to find: F = ? – magnitude of breaking force, t = ? – the time required to stop, We write down formulas which involved the unknown quantities, F = a = FL,’t (2) Some explanations: Formula (1) is simply Newton’s Second Law of Motion, ma formula (2) – the speed decreases from FL too during time t.
Assuming constant breaking force means constant acceleration (deceleration or acceleration directed opposite to direction of motion in this problem), and (2) is the definition of such acceleration. We have three equations with three unknown – that is what algebra requires. From (2) t=FL/a (4) substituting (4) into (3) we get and after a little algebra we get The only unknown in (6) is acceleration a, (7) Substituting (7) to (1) we get And the question (3) from our problem.From data given in the problem we see that (10) Using formula (6) for distance required to stop the car, we have (11) The ratio of distances required to stop the car traveling at these two speeds is (12) And this is the answer to question (3). Substituting the numbers and changing all units to SSL system gives F = 5787 N t = 2.
88 s If you got different numbers you probably forget to change kilometers per hour to meters per second. Reference : http://www. Physics-tutorial.
Net/MM-UP-force-braking-car. HTML Tension in an elevator cable Physics problem An elevator has a mass of keg.What is the tension in the supporting cable when the elevator traveling down at 10 m/s is brought to rest in a distance of 40 m. Assume constant acceleration. Given: m =1400 keg – mass of elevator, v = mm/s – initial speed of the elevator, D = 40 m – distance required to stop the elevator. G = 9.
81 m/so – gravitational acceleration, as usual is assumed to be known. Unknown: T = ? – magnitude of tension in the cable while bringing the elevator to rest. To find T we must calculate: a = ? – acceleration while stopping the elevator, t = ? – time required to stop elevator.Solution. It is convenient to draw a free-body diagram, as in Figure below.
Is the tension in the cable of the elevator, is the gravity force. The resultant force is he force producing acceleration (deceleration in this case) of our elevator. This can be written in the form of the equation if we chose the upward direction as positive. Solving for tension gives (1 a) For further calculations we can drop the vector notation as all the forces are acting along one line.
To calculate the magnitude of the tension T, we must find the magnitude a of the acceleration.It can be found from kinematics equations a = v/t standard formula for distance traveled in motion with constant acceleration (negative in this case as directed opposite to the initial speed). Solving the equations (2) and (3) tit respect to acceleration a, we find (4) Magnitude of tension T can be found from formula (1) taken without the vector notation (magnitude only!! ) Substituting numbers given in the problem we get T = 15484 N. Now it’s time to write equations based on Newton’s Laws. In the vertical direction ml = IN so we can forget about these forces in further analysis.
In the horizontal direction the resultant force exerted on ml is F – FAA and this is the force accelerating block ml . Therefore we can write F – FAA = ml The BFD for mm shows that the only the horizontal force acting on it is the one exerted by block ml . This force has a magnitude of the FAA from the BFD on the left both blocks are in contact they must have the same acceleration a.
So, for the second block the equation of motion is FAA = mm a (2) We can drop out the vector notation from these two equations as the directions are well defined on the Fad’s for both blocks. From the (2) we have a = FAA/mm and substituting this acceleration into (1) we find, after a little elementary algebra, FAA=F ran / +mm) And this is the answer to question (1) from the problem. If the force F is exerted from right to left, as in part (2) of the problem, the analogical seasoning will lead to the answer FEB. = F ml / (ml + mm) Substituting the values given in the problem we get FAA=3.
ON and Feb.=1. AN You can wonder why the force between the blocks is larger when you push from the left. This is because in that situation the block which is a kind of transmitter of force must push the larger mass (mm) than in the second situation, when the larger block is pushing the smaller one. Reference: http://www. Physics-tutorial. Net/MM-UP-moving-blocks.
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