Problem Set #2a



Problem Set #2a

1. Use Vi = 22 m/s, Vf = 0, a = -9/8 m/s2 to calculate time using one of the projectile motion equations. Since the ball was launched from the same height as it returned to, the time to come down is that same.

2. With the given information the problem wants you to assume the Vh is 70. This is done because at small angles of theta, the horizontal component of velocity is very near the resultant angle. For this resultant velocity (70 m/s) and a small angle, the horizontal velocity is probably around 69 m/s or even closer to 70 m/s. So using 70 m/s for Vh and 50 m for Dh, you can calculate the time it took for the arrow to get to the target. Using this time, a Dv of 0 (since it took off and landed at the same height), and acceleration you can calculate Vvinitial. Now knowing Vh and Vvi you can calculate the approximate angle of release.

3. The problem tells you to “consider the angle of projection that theoretically, will produce maximum horizontal displacement when takeoff and landing levels are equal. This means assume angle of 45(. For an angle of 45(, Vh is equal to Vvi, knowing this, the first step is use the horizontal equation and solve for time. You get time = Dh/Vh. You can then plug this into one of the vertical equations where you know Dv = 0, Accel = -9.8, plug in what you got for time above knowing Dh = 3m. Now Vh cancels with 1 of the Vvi’s, since they are equal and you are left to solve for Vvi. This is the initial vertical velocity and also the initial horizontal velocity.

4. Both will jump the same height. By the time the insect and the person leave the ground, only gravity will effect their jump. If they took off with the same initial vertical velocity, their jump height will be equal.

5. First step is to break down the velocity vector into its horizontal and vertical components using 7.5 m/s and 15(. You can use Vvi, Vvf = 0, and Accel to calculate time to the darts highest point. You can then double this to get total time in the air. Now that you have time, you can use Vh with it in the horizontal equation to calculate Dh.

6. You can use –5.0 for Vvi, 2 s for time, and Accel to calculate Dv.

7. As RHR gets more positive, the ideal angle approaches zero. As RHR gets close to 0, RHR approaches 45(.

8. For a vertical jump, one wants a takeoff angle of 90(, for a long jump one wants a takeoff angle slightly less than 45( because RHR is positive. If RHR was 0 for a long jump, the ideal takeoff angle would be 45(.

9. You would use Vresultant in the long R equation along with the angle. Vvi can be used in any one of the four vertical equations, and Vh is used in the horizontal equation (Dh = Vh * time).

10.) It doesn’t. Acceleration only effects an objects vertical motion, and neglecting air resistance leaves no forces in the horizontal direction to effect motion.

11.) Because looking at the long R equation for horizontal range of a projectile, the velocity term is squared. Making it the most important factor for horizontal range. The athletes would have to slow down to achieve their desired takeoff angle, but this would hurt their jumps more than it would help them.

12.) Because they are designed for the metric system and variables like meters, and meter/second. This means they won’t work for feet, feet/s, etc.

Problem Set #2b

1.) a. There may be other forces present (friction, normal force, etc)

b. The earth.

2.) a. No gravity doesn’t change.

b. No, more information is needed to make this conclusion.

c. Yes, this must be why the velocity is greater for the vertical position.

c. No, if it was the same, their speed in each position would be equal.

3.) Use F = ma. The only forces are the athletes weight and the Rn. These forces are on the left side of the equation and the mass and acceleration are on the right. This equation can be solved for acceleration.

4.) (F = m * a and (T = I * (. Force and Torque are similar, mass and moment of inertia are similar, and acceleration and angular acceleration are similar. The angular components are like force, mass, and acceleration except they have a angular “spin”.

5.) The FBD for a box on a ramp should include the angle of the incline, the weight of the box, along with a normal force and a static friction force. If the box was sliding, the static friction force should be dynamic friction.

6.) a. The FBD for the guy on the left should include a weight, probably two normal forces (under each foot), and a friction force under each foot. Finally a force from the umpire on the guy.

b. The guy on the right is the same as the guy on the left, but the friction forces are opposite and the force from the umpire is opposite.

c. The guy in the middle has a weight, a normal force under each foot, and a friction force under each foot keeping his legs from sliding out (like Bambi on ice!). Finally he should have forces from each of the other two guys.

d. The group as a whole should have 1 weight (all of their weights combined), 6 normal forces (1 under each foot), and 6 friction forces (three each direction).

7.) a. The FBD for the engine should include a weight, a normal force under each wheel, as well as a friction force under each wheel (remember to look at the direction the wheel is turning). Finally, it should include a force from the trailer pulling backwards on the engine.

b. The trailer should have all the force the engine has with the exception of a force from the engine going in the opposite direction the force from the trailor acted on the engine.

8.) The maximum value of static friction is found in the equation F = ( * Rn. The actual or required friction is found using (F = m * a. The maximum value is the maximal possible friction statically between the objects given the ( and Rn.

9.) Use (F = m * a. The only force on the book is friction forces from the paper. The static friction is always greater than the dynamic friction so during the initial stages before the paper moves, the force is the greatest. The mass of the book is not zero, so there must be some acceleration. The acceleration is very small and not noticible.

10.) b) place a heavy object in the trunk to increase the weight on the rear wheels. This increases the Rn and therefore increases friction. More friction may mean enough traction to get out of the trunk.

11.) No, with no friction you could not generate any horizontal force to mover you body. ((F = m * a)

12.) Use F = ( * Rn.

13.) Constant velocity tells you to think dynamic friction. Use a tilted coordinate system (aligned to the ramp). Since you are dealing with dynamic friction, you can calculate actual friction using F = ( * Rn. You must calculate the x component of weight as well as the y component of weight. Use the 30( and 10N. (Fy = 0 allows you to calculate Rn. You have Wy and Rn is the only other force. Now with Rn you can use F = ( * Rn to calculate the amount of dynamic friction possible. Do the sum of the forces in the X direction and include a Force for your push. You don’t know this force, but you know Wx and Friction. Solve for force of your push.

14.) Similar to problem 13. Use a tilted coordinate system again, and calculate Wx and Wy. (Fx = 0 allows you to calculate actual or required friction and (Fy = 0 allows you to calculate Rn.

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download