Fisica

Páginas: 7 (1724 palabras) Publicado: 2 de octubre de 2011

R

R
3-22. A jogger runs 2.0 mi west and then 6.0 mi north. Find the magnitude and direction of the resultant displacement.
6.32 mi ;  = 71.60 N of W

50 km/h
20 km/h
R
50 km/h
20 km/h
R
* 3-23. A river flows south with a velocity of 20 km/h. A boat has a maximum speed of 50 km/h in still water. In the river, at maximum throttle, the boat heads due west.What is the resultant speed and direction of the boat?
R = 53.9 km/h, 21.80 S of E
R = 53.9 km/h, 21.80 S of E


300
Fx
F
300
Fx
F
* 3-24. A rope, making an angle of 300 with the horizontal, drags a crate along the floor. What must be the tension in the rope, if a horizontal force of 40 lb is required to drag the crate?
Fx = F cos 300; F = 46.2 N

F
80 N
340
F80 N
340
* 3-25. An vertical lift of 80 N is needed to lift a window. A long pole is used to lift the window. What force must be exerted along the pole if it makes an angle of 340 with the wall?
Fy = F sin 300; F = 96.5 N

B
400 N
200 N
300
B
400 N
200 N
300
* 3-26. The resultant of two forces A and B is 400 N at 2100. If force A is 200 N at 2700, what are themagnitude and direction of force B? ( = 210 - 1800 = 300)
B = -400 N cos 300 = -346 N: B = 346 N, 1800

The Component Method of Vector Addition
R
C
B
A

R
C
B
A

3-27. Find the resultant of the following perpendicular forces: (a) 400 N, 00; (b) 820 N, 2700; and (b) 500 N, 900. Draw each vector, then find R: Ax = +400 N; Bx = 0; Cx = 0: Rx = +400 N
Ay =0; By = -820 N; Cy = +500 N; Ry = 0 – 820 N + 500 N = -320 N
; R = 512 N, 38.70 S of E
3-28. Four ropes, all at right angles to each other, pull on a ring. The forces are A = 40 lb, E; B = 80 lb, N; C = 70 lb, W; and D = 20 lb, S. Find the resultant force on the ring.
A = 40 lb, E
A = 40 lb, E
Ax = +40 lb; Bx = 0; Cx = -70 lb Dx = 0:Rx = +40 lb – 70 lb = -30 lb
Ay = 0; By = +80 lb; Cy = 0; Dy = -20 lb ;
Ry = 0 + 80 lb - 20 lb = +60 lb
; R = 67.1 N, 116.60
600
B
A
600
B
A
*3-29. Two forces act on the car in Fig. 3-27. Force A is 120 N, west and force B is 200 N at 600 N of W. What are the magnitude and direction of the resultant force on the car?Ax = -120; Bx = - (200 N) cos 600 = -100 N
Rx = – 120 N - 100 N; Rx = -220 N
Ay = 0, By = 200 sin 600 = +173 N; Ry = 0 + 173 N = 173 N;
Thus, Rx = -100 N, Ry = +173 N and
Resultant direction: ; R = 280 N, 141.80
A
A
B
B
600
600
*3-30. Suppose the direction of force B in Problem 3-29 is reversed (+1800) and other parameters are unchanged.What is the new resultant? (This result is the vector difference A – B).
The vector B will now be 600 S of E instead of N of E.
Ax = -120 N; Bx = +(200 N) cos 600 = +100 N
Rx = – 120 N + 100 N; Rx = -20 N
Ay = 0, By = -200 sin 600 = -173 N; Ry = 0 - 173 N = -173 N;
Thus, Rx = -20 N, Ry = -173 N and
600
200
A = 60 lb
B = 40 lb
C = 50 lb
600
200
A = 60 lb
B = 40lb
C = 50 lb
Resultant direction: ; R = 174 N, 253.40
*3-31. Determine the resultant force on the bolt in Fig. 3-28. ( Ax = 0 )
Bx = -40 cos 200 = -37.6 lb; Bx = -50 cos 600 = -25.0 lb
Rx = 0 – 37.6 lb – 25.0 lb; Rx = -62.6 lb
Ay = +60 lb; By = 40 sin 200 = 13.7 lb; Cy = 50 sin 60 = -43.3 lb
Ry = 60 lb – 13.7 lb – 43.3 lb; Ry = +30.4 lbR = 69.6 lb  = 25.90 N of W
700
300
300
C
B
A
700
300
300
C
B
A
*3-32. Determine the resultant of the following forces by the component method of vector addition: A = (200 N, 300); B = (300 N, 3300; and C = (400 N, 2500).
Ax = 200 cos 300 = 173 N; Bx = 300 cos 300 = 260 N
Cx = -400 cos 700 = -137 N; Rx = Fx = 296 N
Ay = 200 sin 300 = 100 N;...
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