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Physics 101- Spring 2007 Dynamics Problems Solutions You are helping with set design of the latest play on campus. In the script there is a scene, which calls for the male lead to rise into the air, as if ascending into the clouds. Your plan involves attaching a rope to a harness that the actor will wear. The rope is then put over a pulley and a counterweight is attached to the other end. When the weight is released the actor will be pulled upward. The director has told you that he wants the actor to ascend the 20 feet in 3 seconds. As part of your planning you weigh the actor and find that he has mass of 70kg. You first make some calculations to determine how much concrete to buy (for the counterweight) and which rope to buy. (Your production is on a tight budget and the stronger rope costs more than the thinner, weaker rope. You need to optimize your budget by buying the cheapest rope that will not result in the actor plummeting to the stage.) • Essentially, this is a Newtonian problem with constant forces and acceleration. We will use N2 and N3 to relate the forces to the motion of the actor. • Assume that the actor doesn’t jump off of the stage (no initial velocity). • While a real rope isn’t massless, it will likely have a much smaller mass than the person or block, so assuming an ideal rope isn’t too crazy of an idea. Also, an ideal rope is rigid- does not stretch. • (The pulley should also be treated as having no mass, and being free of friction. This may not be obvious now since we haven’t covered rotation in depth, but it is a necessary assumption. Otherwise, we would need to know the pulley’s mass and the friction within its bearings.) • I would expect the block’s mass to be just a bit more than the actor’s. If it is hundreds of kg’s the actor would feel a very large jerk upwards. If the block’s mass is actually less than the actor’s then, it simply wouldn’t pull the actor up. Let’s assign some variables to the various quantities in the problem (This will keep the mat neat & tidy.) • m= 70kg (actor’s mass) • y= 20’= 6.1 m (distance that the actor has to travel) • t= 3 s (time for the “trip”) • T (tension in the rope) • M (mass of the concrete block)Looking at the motion of the actor we can see that there must be a net force directed upward to provide the force. In the FBD we see that this means that the tension force is greater than the weight. ! F = ma"T # mg = m2yt2$ % & ' ( ) T = m2yt2+ g$ % & ' ( ) This quickly gives us the required tension in the rope, 780N= 175pounds. This means we must buy a rope which is rated for at least 175 pounds. (A wise move would be to spend a few dollars more and get something that gives us some room for error. Better to do that than pay for a trip to the emergency room!) To determine how much concrete we need, we need to look at the FBD of the block, and apply Newton’s Second Law to it as well. ! F = Mab"T # Mg = Mab We know that the tension is the same here as it was on the other side of the pulley, thanks to our simplifications. What is the block’s acceleration? It is the same as the actor’s. (Almost.) Since the rope is not stretching, the two move with the same velocity and acceleration. (For every meter the actor moves up, the block moves down 1 meter.) Hey, what’s with this “almost” business? While the accelerations are the same magnitude, they are in different directions. The block is accelerating downward. ! F = Ma"T # Mg = M#2yt2$ % & ' ( ) m2yt2+ g$ % & ' ( ) # Mg = M#2yt2$ % & ' ( ) T mg T Mg! m2yt2+ g" # $ % & ' = M(2yt2+ g" # $ % & ' M =m2yt2+ g" # $ % & ' (2yt2+ g So, the block’s mass must be 92kg. Does this make sense? Yeah. It is bigger than the actor’s, but not by a whole lot.Over the break you head up to San Francisco to visit a friend from your dorm. Since this is the first time you’ve visited this friend, you find yourself constant looking down at the map which she emailed you. (You might guess where this is going.) As you are driving up a hill, you glance down at the map for just a moment to check the house number when a kid runs out in the street. You look up just in time to slam on your brakes and skid to a stop without harming the kid. While the kid goes on his way, as if he doesn’t even notice you, a police officer did notice you and your 30-foot skid mark. He promptly marches over and writes a ticket for speeding in a 25 mph zone. While you weren’t watching your speedometer you didn’t think you were speeding. Perhaps physics can come to the rescue. You determine that the street makes an angle of 20° with the horizontal. Your car manual tells you that the mass of your car is 1430 kg. Will you fight the ticket in court? • This is another problem where (with our assumptions) we will be able to use Newton’s Laws to relate the motion to the force. • We’re going to assume a constant force on the car. (Constant acceleration.) • We’ll use the simplified model of friction, where the frictional force is simply equal to the coefficient of friction times the normal force. Here we will use the kinetic coefficient, since the tires are sliding on the road. • We’ll simply treat the car as a block and ignore the details of its structure. • We would expect a speed of tens of mph (a typical car speed). • d= 30’= 9.1m (length of your skid mark- which is also the distance over which the frictional force is exerted on the car) • θ=20° (slope of the hill) • m=1430kg (mass of the car) • v (speed of the car just before the skidding began) • µ= 0.8 (this is the kinetic coefficient of friction for rubber on concrete, from our text) • f (frictional force) • N (normal force) First, let’s look at the motion of the car. What is the acceleration? Let’s use one of the kinematic equations to related the distance, acceleration, and speeds (initial and final). ! vf2= vi2+ 2ad0 = v2+ 2ada ="v22dFor now, let’s just set this aside, and look at our FBD, so we can put together Newton’s Second Law. There is actually an issue here- what coordinate system should we use? We can either pick one that is parallel and perpendicular to the horizontal or to the hill. It turns out the only …


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LMU PSY 101 - Dynamics

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