A rubber ball bounces. We'd like to understand how the ball bounces. c. Draw a free-body diagram of the ball during its contact with the ground. Is there a net force acting on the ball? If so, in which direction?

Video transcript

Hey, everyone. So this problem is asking us about free body diagrams. A mechanic lifts a pneumatic tire and releases it in the air. Explain why the tire bounces upward by answering these two questions. One, we're asked to provide a free body diagram of the tire when in contact with the ground. And two, we're asked to specify the direction of the net force acting on the tire if any. So we're gonna, we're going to talk through our multiple choice answers here. And in each of these answers, we have a positive normal force. So that's a normal force acting in the positive Y direction. And then we have a weight force in acting in the negative Y direction. So answer choice. A part one is the normal force and the weight force acting in opposite directions. But of equal magnitude as shown by equal lengths uh of the arrow in this free body diagram and then answer A part B is no net force. Answer B part one is the normal force and the weight acting in opposite directions. But of equal magnitude. Answer B part two is downward answer C part one is the normal force acting in the opposite direction of the weight. But the normal force's magnitude is larger than the weights magnitude as denoted by or as shown by a longer arrow for the normal force in the positive Y direction. And a shorter arrow for the weight in the negative Y direction. And answer C part two is downward answer D part one is the normal force and the positive Y, the weight and the negative Y where the normal force has a larger magnitude than the weight. Answer D part two is no net force. Answer E we have the normal force and the positive Y that's a larger magnitude than the weight force and the negative Y and answer choice E part two is upward. And lastly, we have answer choice F where we have the normal force and the weight acting in opposite directions of equal magnitude. And uh answer F part two is upward. So if we were to think about this problem, when the tire, at the moment that the tire comes in contact with the ground for it to bounce, which means for it to change direction and then move upward, that normal force acting in the upward direction must be larger than the weight acting in the downward or negative direction. And so from there, we can eliminate the answer choice is where the normal force and the weight are of equal magnitude. So that eliminates answer choice A B and F. So now we know, part one, for the rest of these choices is correct. We have the normal force is larger than the weight that results in a net upward force. So if you were to sum the forces, the, the net force would be upward because the normal force magnitude is larger than the weight magnitude. And so that means the correct answer is e where we have the normal force larger than the weight and our net force is in the upward direction. So that's all we have for this one. We'll see you in the next video.