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APPLICATIONS. BASIC RESULTANT PROBLEMS First let’s recall some basic facts from geometry about parallelograms opposite sides are parallel and equal opposite.

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Presentation on theme: "APPLICATIONS. BASIC RESULTANT PROBLEMS First let’s recall some basic facts from geometry about parallelograms opposite sides are parallel and equal opposite."— Presentation transcript:

1 APPLICATIONS

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

3 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? 280 lb 250 lb Let’s take these forces as vectors and make a parallelogram and use its properties. 60°

4 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? 280 lb 250 lb 60° v You can make a parallelogram and the diagonal will be the resultant force showing which direction and how strong the resultant force will be.

5 280 lb 250 lb 60° v Using geometry, can you figure out any of the angles or sides of the lower triangle formed with the vector v as one side? 250 lb 120 ° Opposite sides are equal and the two angles add to 180° 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.  How can we then find  as shown?

6 280 lb 250 lb 60° x y v  We could also solve this problem by making a coordinate system with the forces as vectors with initial point at the origin. 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. The force required to move the object is in a direction parallel to the ramp. If we make the ramp steeper, will either of these forces change? 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 would need to be greater. If we make the ramp even steeper what would have to change? Let’s look at the resultant force in each case. Can you see what happens to the resultant force as the ramp gets steeper?

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°. First look at this triangle and find the missing angle. 60° 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. 30° 250 lb Okay---can you see how to use trig to find the magnitude of the force vector to pull the turtle up the ramp?

10 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 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. from southwest to northeast 45°

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

13 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. x y 350 50 This is the northwest quadrant so wind would blow towards southeast. 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. v u 315° c c = v + u ground speed = = 316.6 mph

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


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