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Tricky test 2 elevator
Tricky test 2 elevator














Repeat the steps above with a different mass. The SI unit of force is the newton, which is based on the kilogram and the meter per second squared. Lab cart on the left, lead weight on the right. Why make two diagrams when you can make one? This level of detail is not necessary for your own personal work, but it is a good idea for me to do it so that my work is less ambiguous to you. Sometimes we're dealing with the lab cart (identified by a subscripted 1), sometimes we're dealing with the lead weight (identified by a subscripted 2), and sometimes we're dealing with the whole system - the cart and weight connected by a string (identified by the lack of a subscript). The challenge in this problem is keeping track of the different objects. To maintain the same acceleration, thrust would have to increase. Now that there is drag, so the thrust has to accelerate the bird and overcome drag. When there was no drag, the thrust accelerated the bird.Since weight does not change, lift does not change. Neither of these quantities is affected by drag, therefore net force does not change. The net force is determined by mass and acceleration.Adding drag does not change the mass of the bird or its location, therefore weight does not change. The value of the gravitational field is determined by the location. Weight is determined by mass and gravity.Adding drag to the problem does not change these measurements, therefore acceleration does not change. Acceleration was computed from measured values of velocity and time.What if air resistance was not negligible? How would this change the values calculated above? How would this change the… The forward force of thrust is the only force acting horizontally, which makes it the net force. According to our diagram, this means lift equals weight. Thus, whatever forces act up must be balanced by forces acting down. There is no net force in the vertical direction. Net force and acceleration always have the same direction. (Down is defined as the direction things move when allowed to fall freely.) W = mg Weight is mass times gravity (the value of the gravitational field on Earth). In this problem, acceleration is computed from its definition. If there was drag, it would point opposite the direction of the bird's motion (in other words, backward).ĭetermine the following quantities for the goose in flight… An object accelerating forward must have some force pushing it forward. A winged object like a bird needs some force to keep it aloft. The mass is still 4.5 kg and the bird still accelerates from rest to 6.0 m/s in 2.0 s.Īll objects have weight. All the measurements given in the problem are still valid for part c of this problem.If we now admit that air resistance was present to some extent, how will this change the computed values of… We just decided to ignore it temporarily. Any object moving through the air will experience air resistance.the magnitude of the forward thrust provided by its wings.the magnitude of the upward lift provided by its wings.the magnitude and direction of the net force acting on it.Determine the following quantities for the goose in flight….

Tricky test 2 elevator free#

Draw a free body diagram of the goose in flight.After 2.0 s of horizontal flight the bird has reached a speed of 6.0 m/s (fast enough to stay aloft, but not so fast that we need to worry about air resistance… at first). It starts from rest on the ground, but after a single step it is completely airborne. The speed of the elevator is decreasing since the acceleration is opposite the velocity.Ī 4.5 kg Canada goose is about to take flight. Net force and acceleration are always in the same direction, since the math says so. The mass of the cheeseburger was given in the problem and we just computed the net force a moment ago. Use Newton's second law of motion to determine the acceleration. The difference is negative, which means the net force is downward. Weight was computed in the previous part of this problem. The normal force is what the scale reads. I think I will let up be the positive direction for this problem. This means the net force is the difference of the two forces. There are only two forces on the cheeseburger and they are opposite each other. Assume the elevator is near the surface of the Earth where gravity is around its standard value. Label the upward pointing arrow "normal" and the downward pointing arrow "weight". Try to make the upward pointing arrow look smaller than the downward one. Draw a box with one arrow pointing up and another pointing down. Normal points up, since the problem didn't say anything about the scale not being level. Weight points down, since it always does. Objects resting on solid surfaces also experience a normal force.














Tricky test 2 elevator