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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 add to 180° These 2 angles add to 180° parallel and equal alternate interior angles are congruent (equal sizes) Let’s take these forces as vectors and make a parallelogram and use its properties. 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? You can make a parallelogram and the diagonal will be the resultant force showing which direction and how strong the resultant force will be. 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? 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° 28.1 How can we then find as shown? v 60° 120° 280 lb sin sin120 250 v 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. v 280 250 2 280 250 cos120 2 2 2 v 459.2 y 250cos60, 250sin 60 We could also solve this problem by making a coordinate system with the forces as vectors with initial point at the origin. v 60° 280 lb 280,0 x We then put the vectors in component form and add them. 280,0 250cos60, 250sin 60 v 280 250cos 60 250sin 60 2 2 459.2 250sin 60 tan 28.1 280 250 cos 60 1 PROBLEMS 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 does is acting NOTon change. the object The so thedue force weight to gravity of the stays object is a force. Gravity vertical and of pulls the same down so the force vectorso magnitude for the the gravity weightvector of the object remains the is same. always vertical. Can you see what happens to the resultant force as the ramp gets steeper? The force required to move the object would is in a direction need to be parallel to the ramp. greater. If we make the ramp steeper, even steeper will either what of would theseto have forces change? change? Let’s look at the resultant force in each case. 60° 250 30° lb 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. 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? opp F sin 30 hyp 250 F 250sin30 125 lb. 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. 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. 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 direction from due north. 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. v 350,0 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 y northwest quadrant so wind would blow towards southeast. 315° v 50 350 u c x u 50cos315,50sin 315 c = v + u 350,0 50cos315,50sin 315 50sin 315 1 ground speed = c = 316.6 mph tan 6.4 350 50 cos 315 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 350,0 v 350 c u 50 x 5.2° u 50cos315,50sin 315 c = v + u 350,0 50cos315,50sin 315 50sin 315 1 ground speed = c = 316.6 mph tan 6.4 350 50 cos 315