1206 - Concepts of Physics Wednesday, October 14th
Demonstrations he spinning chair, etc. hank you Mark! Remember the ice skater example? An ice skater is spinning with both arms and a leg outstretched (a) and then she pulls her arms and leg inward (b). Does her angular velocity increase or decrease?
Notes Midterm: November 02nd 16:30-18:00 For those who have class: 19:30-21:00 (we meet at my office F511) If you still have conflict, you need to come see me in person today or Friday!!!
Notes Friday, we will go over the rules for the midterm Friday, we will also go over the topics for the midterm Friday, we will go over the formula sheet for the midterm
A little more about forces Accelerating blocks. Block 1 (mass m1 = 8.00 kg) is moving on a frictionless 30 incline. his block is connected to block 2 (mass m2 = 22.0 kg) by a massless cord that passes over a massless and frictionless pulley (see figure below - left). Find the acceleration of each block and the tensions in the cord. m1 FN 30.0 W1 30 a m2 a W1 sin30 Free body diagrams - separate incline and vertical motion +y +y +x W1 FN W1 cos30 30 W2 +x W2 Block 1 Block 2 Since both blocks accelerate, there must be a net force acting on each one. he key to solving is to realize that Newton s second law can be used separately for each block to relate the net force to the acceleration. Note also that both blocks have accelerations of the same magnitude a, since they move as a unit. We assume that block 1 accelerates up the incline and choose this direction to be the +x axis. If block 1 in reality accelerates down the incline, then the value obtained for the acceleration will be a negative number.
Let s look at block 1 first. hree forces act on the block: (a) W1 is the weight of block 1 (W1 = m1 * g = 8.00 kg * 9.80 m/s 2 = 78.4 N) (b) is the force applied due to the tension in the cord (c) FN is the normal force that the incline exerts W1 sin30 Free body diagrams - separate incline and vertical motion +y +y +x W1 FN W1 cos30 30 W2 Block 1 Block 2 +x he Weight is the only force that does not point along the x or y axes and its x and y components are given in the diagram. Applying Newton s second law (ΣFx = m1 ax ): ΣFx = -W1 sin30 + = m1 a his equation cannot be solved directly, since both and a are unknown. So, we have to consider block 2 to complete our solution. wo forces act on block 2: (a) W2 is the weight of block 2 W2 = m2 * g = 22.0 kg * 9.80 m/s 2 = 216 N) (b) is exerted as a result of block 1 pulling back on the connecting cord. Since the cord and the pulley are frictionless, the magnitude of and are the same. Applying Newton s second law (ΣFy = m2 ay) to block 2 reveals: ΣFy = -W2 + = - m2 a Now we have two equations with two unknowns and can solve the system. Starting from the second equation we obtain: = W2 - m2 a (*) Substitute in equation above: -W1 sin30 + W2 - m2 a = m1 a And find a as: a = (-W1 sin30 + W2)/(m1 + m2) = (-78.4 N * 0.5 + 216 N)/30 kg = 5.89 m/s 2 Now we know a and can put it back into equation (*): = W2 - m2 a = 216 N - 22.0 kg * 5.89 m/s 2 = 86.4 N
Example: hoisting scaffold A window washer on a scaffold is hoisting the scaffold up the side of a building by pulling downward on a rope, as shown in the figure. he magnitude of the pulling force is 540 N, and the combined mass of the worker and the scaffold is 155 kg. Find the upward acceleration of the unit. P W = mg +y +x W = mg he worker and the scaffold form a single unit, on which the rope exerts a force in three places. he left end of the rope exerts an upward force on the worker s hands. his force arises because he pulls downward with a 540 N force, and the rope exerts an oppositely directed force of equal magnitude on him, in accord with Newton s third law. hus the magnitude of the upward force is 540 N and is the magnitude of the tension of the rope. We assume the masses of the pulley and each rope are negligible and the pulleys are frictionless. herefore a 540 N tension force acts upward on the right side of the scaffold pulley. A tension force is also applied to the point P, where the rope attaches to the roof. he roof pulls back on the rope in accord with the third law, and this pull leads to the 540 N tension force that acts on the left side of the scaffold pulley. In addition the weight of the unit must be taken into account: W = mg = 155 kg * 9.80 m/s 2 Newton s second law can be applied to find a: ΣFy = m * ay ΣFy = + + - W = m ay ay = (3 - W)/m = (3*540 N - 155 kg * 9.80 m/s 2 )/155 kg = 0.65 m/s 2
Pulleys - a few more things Pulleys are used on many occasions - usually to make the necessary force to lift something smaller. A Pulley is a simple machine made with a rope, belt or chain wrapped around a wheel. A pulley changes the direction of a force. here are three kinds of pulleys: (a) a fixed pulley (b) a movable pulley (c) a combined pulley A single pulley changes the direction of the lifting force. For example, if you are lifting a heavy object with a single pulley anchored to the ceiling, you can pull down on the rope to lift the object instead of pushing up. he same amount of effort is needed as without a pulley, but it feels easier because you are pulling down. A movable pulley is a pulley that moves with the load. he movable pulley allows the effort to be less than the weight of the load. he movable pulley also acts as a second class lever. he load is between the fulcrum and the effort. If you add a second pulley, the amount of effort to lift the heavy object is much less. For example, to lift a box weighing 150 N, one would need to exert 150 N of force without the help of pulleys. However, by using just two pulleys, the person would only need to use 50 N of force. A combined pulley (one fixed and one movable pulley) makes life easier as the effort needed to lift the load is less than half the weight of the load.
his figure shows a single pulley with a weight on one end of the rope. he other end is held by a person who must apply a force to keep the weight hanging in the air (in equilibrium). here is a force (tension) on the rope that is equal to the weight of the object. his force or tension is the same all along the rope. In order for the weight and pulley (the system) to remain in equilibrium, the person holding the end of the rope must pull down with a force that is equal in magnitude to the tension in the rope. For this pulley system, the force is equal to the weight, as shown in the picture. he mechanical advantage of this system is 1! In the second figure, the pulley is moveable. As the rope is pulled up, it can also move up. he weight is attached to this moveable pulley. Now the weight is supported by both the rope end attached to the upper bar and the end held by the person! Each side of the rope is supporting the weight, so each side carries only half the weight (2 upward tensions are equal and opposite to the downward weight, so each tension is equal to 1/2 the weight). So the force needed to hold up the pulley in this example is 1/2 the weight! he mechanical advantage of this system is 2; it is the weight (output force) divided by 1/2 the weight (input force).
he mechanical advantage of each system is easy to determine. Count the number of rope/ cable segments on each side of the pulleys, including the free end. If the free end is to be pulled down, subtract 1 from this number. his number is the mechanical advantage of the system! o compute the amount of force necessary to hold the weight in equilibrium, divide the weight by the mechanical advantage! Here there are 3 sections of rope. Since the applied force is downward, we subtract 1 for a mechanical advantage of 2. It will take a force equal to 1/2 the weight to hold the weight steady. his figure has the same two pulleys, but the rope is applied differently and it is pulled upwards. he mechanical advantage is 3, and the force to hold the weight in equilibrium is 1/3 the weight.
YOUR URN: When designing the system you will utilize to raise and lower a V, etc., think about the load you plan to lift. 1. How heavy is it? 2. Can pulley positions be placed to take advantage of leverage? Particularly important to remember is that both sides of your platform MUS be raised with identical pulley layouts. If this isn't done, one side will move faster than the other --
ypical pulley problem for assignment (and midterm) his here is the simplest system - it can get far more complex. But the physics doesn t change... his is a pulley system where masses m and M are connected by a rope over a massless and frictionless pulley. Note that M > m and both masses are at the same height above the ground. he system is initially held at rest, and is then released. Calculate the acceleration of the masses, the velocity of mass m when it moves a distance h, and the work done by the tension force on mass m as it moves a distance h. Before we start calculating, let s go through the three steps for problem solving: 1. Ask yourself how the system will move 2. Choose a coordinate system 3. Draw free-body diagrams YOUR URN
mg Mg +y 1. Ask yourself how the system will move: From experience, we know that the heavy mass, M, will fall, lifting the smaller mass, m. Because the masses are connected, we know that the velocity of mass m is equal in magnitude to the velocity of mass M, but opposite in direction. Likewise, the acceleration of mass m is equal in magnitude to the acceleration of mass M, but opposite in direction. 2. Choose a coordinate system: You should choose the coordinate system that it simplifies your calculation. In this case, let s follow the standard convention of saying that up is the positive y direction and down is the negative y direction. 3. Draw free-body diagrams: We know that this pulley system will accelerate when released, so we shouldn t expect the net forces acting on the bodies in the system to be zero. 1. What is the acceleration of mass M? YOUR URN
mg Mg +y 1. What is the acceleration of mass M? Because the acceleration of the rope is of the same magnitude at every point in the rope, the acceleration of the two masses will also be of equal magnitude. If we label the acceleration of mass m as a, then the acceleration of mass M is a. Using Newton s Second Law we find: for mass m: - mg = ma for mass M: - Mg = -Ma By subtracting the first equation from the second, we find (M m)g = (M + m)a or a = (M m)g/(m + m). Because M m > 0, a is positive and mass m accelerates upward as anticipated. his result gives us a general formula for the acceleration of any pulley system with unequal masses, M and m. Remember, the acceleration is positive for m and negative for M, since m is moving up and M is going down. g (M - m) a = (M + m) 2. What is the velocity of mass m after it travelled a distance h? YOUR URN
mg Mg +y 2. What is the velocity of mass m after it travelled a distance h? We can tackle this problem in terms of energy. Because the masses in the pulley system are moving up and down, their movement corresponds to a change in gravitational potential energy. Because mechanical energy E is conserved, we know that any change in the potential energy PE of the system will be accompanied by an equal but opposite change in the kinetic energy KE of the system: ΔKE = -ΔPE Remember that since the system begins at rest, KE0=0. As the masses move, mass M loses Mgh joules of potential energy, whereas mass m gains mgh joules of potential energy. Applying the law of conservation of mechanical energy, we find: 1/2Mv 2 + 1/2mv 2 = -(mgh - Mgh) --> 1/2(M+m)v2 = gh(m-m) v 2 = 2gh (M-m)/(M+m) Mass m is moving in the positive y direction. 3. What is the work done by the force of tension in lifting mass m a distance h? YOUR URN
mg Mg +y 3. What is the work done by the force of tension in lifting mass m a distance h? Since the tension force is in the same direction as the displacement h we know that the work done is equal to h. But what is the magnitude of the tension force? We know that the sum of forces acting on m is mg which is equal to ma. herefore, = m(g a). From the solution to question 1, we know that a = g(m m)/(m + m), so substituting in for a, we get: W = h = h m(g-a) = mgh - mha = mgh - mh[g(m-m)/(m+m)] W = mgh (1 - ) M-m M+m
Many examples... Pulleys are used everywhere: industry, climbing, sailing, etc.