In school tasks in physics on determination of friction force of sliding the rectilinear uniform or rectilinear uniformly accelerated motion of a body generally is considered. Look how it is possible to find friction force in different cases of dependence on statements of the problem. Correctly to estimate influence of forces and to work out the equation of the movement, always draw the drawing.
Instruction
1. Case 1. A formula for sliding friction force: Ftr = mn, where m – coefficient of a sliding friction, N – force of reaction of a support, N. For the body sliding on the horizontal plane, N = G = mg where G is body weight, N; m – body weight, kg; g – acceleration of gravity, m / с2. Values of dimensionless coefficient of m for this couple of materials are given in reference books. Knowing the body weight and couple of materials. sliding from each other, find friction force.
2. Case 2. Consider the body sliding on a horizontal surface and moving uniformly accelerated. It is affected by four forces: force, the setting body in motion, the gravity, force of reaction of a support, sliding friction force. As a surface horizontal, force of reaction of a support and gravity are directed along one straight line and counterbalance each other. Movement describes the equation: Fdv - Ftr = ma; where Fdv is the module of force setting a body in motion, N; Ftr is the module of friction force, N; m – body weight, kg; a – acceleration, m / с2. Knowing values of weight, acceleration of a body and force influencing it find friction force. If these values are not set directly, look whether there are in a condition data from which it is possible to find these sizes.
3. Example of a task 1: on the bar weighing 5 kg lying on a surface use force of 10 N. As a result bar moves uniformly accelerated and there pass 10 meters in 10 seconds. Find sliding friction force. The equation for the movement of bar: Fdv - Ftr = ma. The way of a body for the uniformly accelerated movement is set by equality: S = 1/2at^2. From here you can define acceleration: a = 2S/t^2. Substitute these conditions: and = 2*10/10^2 = 0.2 m / с2. Now find equally effective two forces: ma = 5*0.2 = to 1 N. Vychislita friction force: Ftr = 10-1 = 9 N.
4. Case 3. If the body on a horizontal surface is at rest, or moves evenly, under the second law of Newton of force are in balance: Ftr = Fdv.
5. Example of a task 2: reported to the bar weighing 1 kg which is on a plain surface an impulse as a result of which it passed 10 meters in 5 seconds and stopped. Determine sliding friction force. As well as in the first example, sliding of bar is influenced by force of the movement and friction force. As a result of this influence the body stops, i.e. balance comes. Equation of the movement of bar: Ftr = Fdv. Or: N*m = ma. Bar slides uniformly accelerated. Calculate its acceleration like a task 1: a = 2S/t^2. Substitute values of sizes from a condition: and = 2*10/5^2 = 0.8 m / с2. Now find friction force: Ftr = ma = 0.8*1 = 0.8 N.
6. Case 4. The body which is spontaneously sliding on the inclined plane is affected by three forces: gravity (G), force of reaction of a support (N) and friction force (Ftr). Gravity can be written down in such look: G = mg, N where m is body weight, kg; g – acceleration of gravity, m / с2. As these forces are directed not along one straight line, write down the equation of the movement in a vector look.
7. Having put by the rule of a parallelogram of force of N and mg, you receive the resulting F force’. From the drawing it is possible to draw conclusions: N = mg*cosα; F’ = mg*sinα. Where α – a plane tilt angle. Friction force can be written down a formula: Ftr = m*n = м*mg*cosα. The equation for the movement takes a form: F '-Ftr = ma. Or: Ftr = mg*sinα-ma.
8. Case 5. If to a body the additional force of F directed along the inclined plane, then friction force is applied to be expressed: Ftr = mg*sinα+F-ma if the direction of the movement and force of F coincide. Or: Ftr = mg*sinα-F-ma if force of F counteracts the movement.
9. Example of a task 3: bar weighing 1 kg slid off top of the inclined plane in 5 seconds, having passed a way of 10 meters. Determine friction force if plane 45o tilt angle. Consider also a case when bar was influenced by the additional force of 2 N applied along a tilt angle in the direction of the movement. Find acceleration of a body similar to examples 1 and 2: and = 2*10/5^2 = 0.8 m / с2. Calculate friction force in the first case: Ftr = 1*9.8*sin(45о)-1*0.8 = 7.53 N. Opredelita friction force in the second case: Ftr = 1*9.8*sin(45о) +2-1*0.8 = 9.53 N.
10. Case 6. The body moves on an inclined surface evenly. Means, under the second law of Newton the system is in balance. If sliding spontaneous, the movement of a body submits to the equation: mg*sinα = Ftr. If to a body the additional force (F) interfering uniformly accelerated movement is applied, expression for the movement has an appearance: mg*sinα–Fтр-F = 0. From here find friction force: Ftr = mg*sinα-F.