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APPLICATIONS

APPLICATIONS. BASIC RESULTANT PROBLEMS. First let’s recall some basic facts from geometry about parallelograms. opposite sides are. These 2 angles add to 180 °. These 2 angles add to 180 °. These 2 angles are equal. These 2 angles are equal. opposite sides are. parallel and equal.

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APPLICATIONS

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  1. APPLICATIONS

  2. BASIC RESULTANT PROBLEMS First let’s recall some basic facts from geometry about parallelograms opposite sides are These 2 angles add to 180° These 2 angles add to 180° These 2 angles are equal These 2 angles are equal opposite sides are parallel and equal These 2 angles are equal These 2 angles are equal These 2 angles add to 180° These 2 angles add to 180° parallel and equal alternate interior angles are congruent (equal sizes)

  3. Let’s take these forces as vectors and make a parallelogram and use its properties. 250 lb 60° 280 lb Two draft horses are pulling on a tree stump with forces of 250 pounds and 280 pounds as shown. If the angle between the forces is 60°, then what is the magnitude of the resultant force? What is the angle between the resultant and the 280 pound force?

  4. You can make a parallelogram and the diagonal will be the resultant force showing which direction and how strong the resultant force will be. 250 lb v 60° 280 lb Two draft horses are pulling on a tree stump with forces of 250 pounds and 280 pounds as shown. If the angle between the forces is 60°, then what is the magnitude of the resultant force? What is the angle between the resultant and the 280 pound force?

  5. Using geometry, can you figure out any of the angles or sides of the lower triangle formed with the vector v as one side? Opposite sides are equal and the two angles add to 180° How can we then find  as shown? 250 lb v 250 lb 60° 120°  280 lb Looking at the lower triangle we have side, angle, side so we can use the Law of Cosines to find the magnitude (length) of v.

  6. y We could also solve this problem by making a coordinate system with the forces as vectors with initial point at the origin. 250 lb v 60°  x 280 lb We then put the vectors in component form and add them.

  7. INCLINED PLANE PROBLEMS

  8. Think of pushing or pulling something up a ramp. Our model will assume you have a well-oiled dolly and we will neglect friction. What other forces are there in this problem? Gravity is acting on the object so the weight of the object is a force. Gravity pulls down so the force vector for the weight of the object is always vertical. Gravity does NOT change. The force due to gravity stays vertical and of the same magnitude so the gravity vector remains the same. The force required to move the object is in a direction parallel to the ramp. The force required to move the object would need to be greater. If we make the ramp steeper, will either of these forces change? If we make the ramp even steeper what would have to change? Can you see what happens to the resultant force as the ramp gets steeper? Let’s look at the resultant force in each case.

  9. Workers at the zoo must move a 250-pound giant tortoise to his new home. Find the amount of force required to pull him up a ramp leading into a truck if the angle of elevation of the ramp is 30°. 60° 250 lb 30° First look at this triangle and find the missing angle. Now look at this triangle. We can easily find the other angle and the hypotenuse since it is part of a parallelogram and parallel to the 250 lb. side. Okay---can you see how to use trig to find the magnitude of the force vector to pull the turtle up the ramp?

  10. Navigation Problems We can represent the speed and direction of a plane in still air as a vector. We’d need to add to that, the speed and direction of the wind. The resultant vector would be the speed and direction the plane would actually travel.

  11. to northeast 45° from southwest In these problems, they will tell you the direction the wind is coming FROM---NOT the direction it is blowing. For example: If a wind is blowing from the southwest, it is blowing towards the northeast at a 45° angle.

  12. bearing measured clockwise from north In this case, 270° plus the drift angle. In air navigation, the bearing is a nonnegative angle smaller than 360° measured in a clockwise diredtion from due north.

  13. ground speed = = 316.6 mph An airplane heads west at 350 miles per hour in a 50 mph northwest wind. Find the ground speed and bearing of the plane. Let’s draw a picture on coordinate axes. You could draw the parallelogram and use a triangle and trig to find the resultant vector whose magnitude is the groundspeed and use the angle to determine the bearing, or you could put these 2 vectors in component form and add. Let’s do the second way this time. This is the northwest quadrant so wind would blow towards southeast. y 315° v x 50 350 u c c = v + u

  14. ground speed = = 316.6 mph An airplane heads west at 350 miles per hour in a 50 mph northwest wind. Find the ground speed and bearing of the plane. Remember the bearing is measured clockwise from north y bearing = 270° – 6.4° = 263.6° v x 5.2° 50 350 u c c = v + u

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