1. Inclined Plane Problems

2. Axes for Inclined Planes X axis is parallel to the inclined plane Y axis is perpendicular to the inclined plane Friction force (f or F f ) is on the x axis opposite the motion Weight or Force of Gravity (F g ) is straight down and must be split into x and y components. Normal force (F N ) is upward along the y axis Applied force (F a ) may be in either direction on the x axis or at angle (which would require it to be split into x and y components)

3. Forces Acting on the Object FaFa FNFN f F gx FgFg F gy Note: The applied force and the force of friction can be in either direction as long as the friction force is opposite to the motion.   F gx x y Remember: Weight or force of gravity = F g = mg (some may use different symbols such as W or F W for weight)

4. Useful Equations for Inclined Plane Problems Formulas F net,x =  F x = ma x F net,y =  F y = ma y F g = mg F gx = mg sin θ F gy = mg cos θ F f =  F N * Assumes that Ɵ is as shown on previous diagram. Explanation of Symbols F net = net force m = mass F g = weight (force of gravity) F f or f = friction force  = coeff. of friction F N = normal force F a = applied force

5. Problems With No Acceleration When the object on the plane does not have any acceleration, then the forces acting on the object in both x and y must be equal to zero. FaFa FNFN f F gx FgFg F gy  F gx x y  F x = F a - f - F gx = ma x = 0  F y = F N - F gy = ma y = 0 Note: There might not be an applied force or it may be down the incline (so f would be up the incline).

6. Problems With Acceleration When the object on the plane has acceleration, there is a net force along the incline (x) but the net force perpendicular to incline (y) is still zero. FaFa FNFN f F gx FgFg F gy  F gx x y  F x = F a - f - F gx = ma x  F y = F N - F gy = ma y =0 Note: There might not be an applied force or it may be down the incline (so f would be up the incline).

7. Tension Problems

8. What is Tension? Tension is defined as a force transmitted along a rope, chain, or wire. Tension will remain constant throughout the length of the rope. Tension is treated as a force in force diagrams and calculations Tension is measured in force units. (Newtons, dynes, or pounds)

9. Weight or mass? What’s the difference? Mass is the amount of matter that an object is made up of, measured in mass units (g, kg, slugs) The mass of an object remains constant at any location in the universe unless matter is added or removed Weight is the force of gravity acting on an object, measured in force units (N, dynes, or pounds) Weight depends on the acceleration due to gravity for the location of the object. Weight (w) = mg

10. Problems in Equilibrium Equilibrium means that the object has a net force and acceleration of zero. Since F net,x or  F x = 0, the x-components of tension should cancel each other. (F right = F left ) Since F net, y or  F y = 0, the y-components of tension should cancel the objects weight. (F up = F down )

11. Force Diagram (Free-Body Diagram) F g = mg T1T1 T2T2 11 22 T 2x =T 2 cos  2 T 2y =T 2 sin  2 T 1y =T 1 sin  1 T 1x =T 1 cos  1 11 22

12. Applying Newton’s 2nd Law θ1θ1 θ2θ2 T1T1 T2T2 Since the sign is in equilibrium, the net force must be equal to zero.  F y = T 1 sin θ 1 +T 2 sin θ 2 – Fg = 0  F y =T 1 cos θ 1 –T 2 cos θ 2 = 0 F g = mg

13. Applying Newton’s Laws to Systems with Multiple Objects.

14. Atwood Machine An atwood machine is a very common system in classical physics It is made of two masses tied together by a string that passes over a pulley Both objects could be hanging and the system would only have motion in the vertical direction OR one object could be sliding along a table top with the other hanging over the side Assumptions made (for now): the pulley is “massless” and “frictionless”. This allows us to solve the problem as if the pulley was not there at all. Tension in the string acts equally on both objects and they both accelerate at the same rate.

15. A typical Atwood machine problem m1m1 m2m2 Two masses, m 1 and m 2, are tied together with a string that is passed over a light, frictionless pulley. Mass m 2 is accelerating downward. If the coefficent of friction between mass m 1 and the table is μ. Find the tension, T, in the string and the acceleration, a, of the masses.

16. To Solve…simply follow the steps we have already established for Force problems. Draw the force diagram –Include all forces acting on each object –Label all forces with a symbol, NOT NUMBERS Define an axis system on each object Write the Newton’s 2 nd Law equations for each object. Plug in what you know and solve the systems of equations for what you don’t know. (Algebra)

17. Setting up the Atwood problem First with no numbers! m2m2 m1m1 FNFN T f F g1 =m 1 g F g2 =m 2 g T a a = + NOTE: When writing the equations for m 2, consider the direction of acceleration to be positive. Write the equations for m1: 0 Write the equations for m 2 : There are no forces along the x-axis for m 2. x y

18. Solving the Atwood Problem Algebra Now that you have written the equations for this problem, it is time to plug in any values that you know and solve the system of equations for Tension and acceleration. Keep in mind that there is more than one algebra technique that can be used, this is an example of substitution. Any properly done method will work. LeLe Substitute values given in the problem. Let… m1 = 4 kg m2 = 2 kg μ= 0.35 For mass m 1 : For mass m 2 : Now solve for a and T

19. Solving the Atwood Problem Algebra (continued) Now that we have plugged in values given in the problem and simplified the equations, we must solve this system of equations for a and T. This example will use substitution. Solve one of the equations for T and plug it into the other.

20. Other objects connected

21. Objects connected m1m1 m2m2 m3m3 Example: Three objects, m 1, m 2, and m 3 are tied together by a rope and pulled along a level surface by an applied force Fa. All three objects have the same coefficient of friction, μ. Find the tensions in the ropes connecting the masses. Some real life situations that this model applies to include: train cars connected by couplings, find the tension in the couplings; pulling sleds tied together; pulling a trailer (only 2 objects)…

22. Objects connected Drawing a force diagram m1m1 m2m2 m3m3 F g1 =m 1 gF g3 =m 3 gF g2 =m 2 g FaFa F N2 F N1 T2T2 T2T2 T1T1 T1T1 f 1 =μF N1 f 2 =μF N2 f 3 =μF N3 F N3

23. Objects connected Write the force equations for each object. a For mass m 1 : For mass m 2 : For mass m 3 : m1m1 m2m2 m3m3 F g1 =m 1 gF g3 =m 3 gF g2 =m 2 g FaFa F N2 F N1 T2T2 T2T2 T1T1 T1T1 f 1 =μF N1 f 2 =μF N2 f 3 =μF N3 F N3

24. Objects connected Solving the equations To solve this problem further we will need to plug in any known values and work the equations from one end to the other. In this problem, the normal force, F N, is equal to the weight of the block which can be shown by solving each of the y-equations for normal force. We will work this problems under two conditions: 1) constant velocity, and 2) accelerated motion. Let m 1 = 10 kg, m 2 = 20 kg, m 3 = 15 kg, F a = 110, μ = 0.25

25. Objects connected Case 1: Constant velocity

26. Objects connected with constant velocity, solving mass 3 Let m 1 = 10 kg, m 2 = 20 kg, m 3 = 15 kg, F a = 110, μ = 0.25 Tension between masses m 2 and m 3.

27. Objects connected constant velocity, solving mass 2 Let m 1 = 10 kg, m 2 = 20 kg, m 3 = 15 kg, F a = 110, μ = 0.25 T 2 = 73.2 N Tension in the string between masses m 1 and m 2.

28. Objects connected constant velocity, summary So we started with a force diagram, then wrote equations for each mass. We did not need to evaluate the equations for the last mass because we had solved for both tensions already, so we were done. Note: We could have started at mass 3 and worked back to mass 1. You could solve this problem for applied force to keep the object moving at constant velocity in this method. Applied force did not need to be given.

29. Objects connected Case 2: Accelerated motion

30. Objects connected Write the force equations for each object. a For mass m 1 : For mass m 2 : For mass m 3 : m1m1 m2m2 m3m3 F g1 =m 1 gF g3 =m 3 gF g2 =m 2 g FaFa F N2 F N1 T2T2 T2T2 T1T1 T1T1 f 1 =μF N1 f 2 =μF N2 f 3 =μF N3 F N3

31. Objects connected accelerated motion, must find acceleration first To find acceleration to use for this problem you can treat the whole system together as one object being accelerated under the influence of the applied force. m 1 + m 2 + m 3 FNFN FaFa F g = (m 1 + m 2 + m 3 )*g f Now we have acceleration of each mass to use as we solve for tensions.

32. Objects connected accelerated motion, solving mass 1 On the last example we started with mass 3 and worked toward mass 1. This time I am beginning at mass 1 and moving toward mass 3 for no other reason than to show that you can work these problems either way. Let m 1 = 10 kg, m 2 = 20 kg, m 3 = 15 kg, F a = 150N, μ = 0.25 a = 0.88 m/s 2

33. Objects connected accelerated motion, solving mass 2 Let m 1 = 10 kg, m 2 = 20 kg, m 3 = 15 kg, F a = 150N, μ = 0.25 a = 0.88 m/s 2 T 1 = 33.3 N