PHYS 106 Spring 2003

"How Things Work" Problem Set #1

(with solutions)

Concise explanations of 1 or 2 sentences for each part are best.

 

ONE. Study the photograph, on page 1 of the text, of the "tablecloth" trick. Notice that the dishes move a little bit during the time the tablecloth slides under them.
  1. What provides the force that pushes the dishes to the right?

    2. friction between the dishes and the tablecloth

  2. How does the inertia of the dishes make this trick work?

    3. The dishes want to remain at rest because of their inertia, and they do almost stay at rest. However, the small frictional force pushes them to the right during the brief time the tablecloth is moving under them.

  3. What would happen if the dishes didn't have any inertia?

Assuming they were still compressed against the tablecloth, so that the friction still existed, but they had no inertia, they would go flying off the table along with the tablecloth.

TWO. Consider what it would be like if you went to the surface of the Moon, as compared to the surface of the Earth where you are now.
  1. Explain, in general, the difference between the weight of an object and its mass.

    2. Weight is a measure of gravitational attraction, to the Earth for example. Mass is a measure of inertia.

  2. Is your weight the same on the Moon as on the Earth? What about your mass?

    3. Mass it the same everywhere. But the Moon has less gravity, so your weight is less there.

     

  3. Explain a simple way to check your weight on the Earth and your weight on the Moon, assuming you could travel to the Moon.

    4. An ordinary bathroom scale would work fine. You will weigh less on the Moon.

  4. Explain a simple way to check and see if your mass is the same on the Earth as on the Moon, again assuming you could travel to the Moon. There is more than one possible answer. Your test should not involve vertical motions, since that brings in gravity.

Any horizontal change in motion would test mass-- for example making a sharp left or right turn and noticing how difficult it is. It will be the same difficulty on the Earth and the Moon. Or push off a wall or a friend while on a low-friction surface.

 

THREE. "Case 2" on page 36 of your text (Diving platform). Do all parts a, b, c, d, e and f. Be careful with part f. "Speed" means total speed in whatever direction you are travelling, not just the horizontal part or just the vertical part.
  1. 1 1/2 seconds is pretty close
  2. 15 m/s downward is pretty close
  3. More. To prove this, notice that when you pass the level of the platform on the way down, you are already moving downward. You then gain speed as you fall to the water.
  4. She will hit the water first. Giving yourself an initial upward velocity will make you hit the water with greater speed, but it will take longer to get there.
  5. You'll hit at the same time.
  6. The person who runs off horizontally will have greater total speed when hitting the water. The vertical portions are the same for each, but the horizontal person also has some horizontal velocity.

FOUR. This is a follow-up to "Case 2" above. If you jump upward as you leave the platform, you will momentarily come to rest at the top of the motion. In what direction is your acceleration at that moment, or is it zero? Explain.

 

Acceleration is always down for this type of motion, even at the top when velocity is zero. If acceleration WERE zero, you'd stay at zero velocity forever once you reached the top!

 

FIVE. "Case 6" on page 37 of your text (Bicycle trip). Do only parts a, b, c, and d. Omit part e.
  1. Net force on you is zero. You are not yet accelerating.
  2. Since you're moving at constant velocity, the net force on you is still zero. In other words, all forces are balanced against each other.
  3. If you lifted the object vertically through 1 meter, you would do (400 N)(1m), which is 400 N-m (or Joules) of work. If you go up the slope, the total work must be the same. The distance is 10 times more, so the force must be one-tenth, which is 40 N of force.
  4. The work is the same by either route, which is (weight)(vertical change in height). This is (400 N)(500m), which is 200,000 Joules.

 

 

SIX. Consider an ordinary seesaw with negligible friction at the pivot. It is a total of 10 meters long, or in other words 5 meters from the pivot to each end. With no children on it, it is perfectly balanced.
  1. Now a 32 kg child sits at the very left end. How far to the right of the pivot must a 40 kg child sit to balance the seesaw? You must show the equation you are using to get your answer.

    2.  

    Balance the total torque (which is force times lever arm) about the pivot. You can omit the 9.8 m/s2 factor from each weight if you wish, since it will cancel out, but I will include it:

    (32 kg)(9.8 m/s2)(5 m) = (40 kg)(9.8 m/s2)(X m)

    Solving for X gives 4 meters.

  2. The 32 kg child remains at the left end, but the 40 kg now moves all the way to the right end, so that the seesaw is unbalanced. A third child, of mass 20 kg, wants to climb aboard. How far to the left of the pivot must this child sit to balance the seesaw? All three children are now on the seesaw. Again, show your equation.

 

There are now two torques on the left side, balancing one on the right side:

(32 kg)(9.8 m/s2)(5 m) + (20 kg)(9.8 m/s2)(X m) = (40 kg)(9.8 m/s2)(5 m)

Solving for X gives 2 meters.

SEVEN. A car is driving on a level road at 51 mph. It is chasing a truck moving at 50 mph. The truck has a gently sloping ramp hanging from the back, and once the car reaches the truck it attempts to drive up the ramp into the truck. Can this safely work, or will the car be moving so fast once it enters the truck that it crashes into the passenger compartment at the front of the truck? Explain.

It will work. We discussed this in class. The car is only moving 1 mph relative to the ramp, so that's how fast it will be moving on the ramp as soon as it touches it. However, the tires of the car are spinning really fast, and if you kept the accelerator pedal depressed, eventually you would be going 51 mph through the truck. But this would take several seconds, giving you plenty of time to pull your foot off the gas, and stop the car inside the truck.

  

EIGHT. "Case 2" on page 75 of your text (Revolving door). Do all parts a, b, c, d, and e.
  1. Since torque is (force)(lever arm), pushing further from the axis gives a larger lever arm and therefore a larger torque.
  2. Applying an unbalanced torque causes an angular acceleration.
  3. Inertia. Once it reaches some angular velocity, the door wants to maintain it, unless a torque acts to change it. There is some friction, but it's pretty small.
  4. Friction. Friction causes a torque to slow the door down.
  5. You want to minimize the "rotational inertia" also known as "moment of inertia". To do this, you want to minimize the amount of mass far from the axis. So don't put too much mass on the outside edge.