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Chapter 4. Translational Equilibrium and Friction.

Note: For all of the problems at the end of this chapter, the rigid booms or struts are considered to be of negligible weight. All forces are considered to be concurrent forces.

Free-body Diagrams

4-1. Draw a free-body diagram for the arrangements shown in Fig. 3-18. Isolate a point where the important forces are acting, and representeach force as a vector. Determine the reference angle and label components.

(a) Free-body Diagram (b) Free-body with rotation of axes to simplify work.
4-2. Study each force acting at the end of the light strut in Fig. 3-19. Draw the appropriate free-body diagram.
There is no particular advantage to rotating axes.
Components should also be labeled on diagram.Solution of Equilibrium Problems:

4-3. Three identical bricks are strung together with cords and hung from a scale that reads a total of 24 N. What is the tension in the cord that supports the lowest brick? What is the tension in the cord between the middle brick and the top brick?
Each brick must weight 8 N. The lowest cord supports only one brick,
whereas the middle cordsupports two bricks. Ans. 8 N, 16 N.
4-4. A single chain supports a pulley whose weight is 40 N. Two identical 80-N weights are then connected with a cord that passes over the pulley. What is the tension in the supporting chain? What is the tension in each cord?
Each cord supports 80 N, but chain supports everything.
T = 2(80 N) + 40 N = 200 N. T = 200 N
*4-5. Ifthe weight of the block in Fig. 4-18a is 80 N, what are the tensions in ropes A and B?
By - W = 0; B sin 400 – 80 N = 0; B = 124.4 N
Bx – A = 0; B cos 400 = A; A = (124.4 N) cos 400
A = 95.3 N; B = 124 N.

*4-6. If rope B in Fig. 4-18a will break for tensions greater than 200 lb, what is the maximum
weight W that can be supported?(Fy = 0; By – W = 0; W = B sin 400; B = 200 N
W = (200 N) sin 400; W = 129 lb


*4-7. If W = 600 N in Fig. 18b, what is the force exerted by the rope on the end of the boom A in Fig. 18b? What is the tension in rope B?
(Fx = 0; A – Wx = 0; A = Wx = W cos 600
A = (600 N) cos 600 = 300 N
(Fy = 0; B – Wy = 0; B = Wy = W sin 600B = (600 N) sin 600 = 520 N
A = 300 N; B = 520 N

*4-8. If the rope B in Fig. 18a will break if its tension exceeds 400 N, what is the maximum
weight W? (Fy = By - W = 0; By = W
B sin 400 = 400 N ; B = 622 N (Fx = 0
Bx – A = 0; B cos 400 = A; A = (622 N) cos 400 A = 477 N.
*4-9. What is the maximum weight W forFig. 18b if the rope can sustain a maximum tension of only 800 N? (Set B = 800 N).
Draw diagram, then rotate x-y axes as shown to right.
(Fy = 0; 800 N – W Sin 600 = 0; W = 924 N.
The compression in the boom is A = 924 Cos 600 A = 462 N.
*4-10. A 70-N block rests on a 300 inclined plane. Determine the normal force and find the friction force that keeps the blockfrom sliding. (Rotate axes as shown.)
(Fx = N – Wx = 0; N = Wx = (70 N) cos 300; N = 60.6 N
(Fx = F – Wy = 0; F = Wy = (70 N) sin 300; F = 35.0 N
*4-11. A wire is stretched between two poles 10 m apart. A sign is attached to the midpoint of the line causing it to sag vertically a distance of 50 cm. If the tension in each line segment is 2000 N, what is the weightof the sign? (h = 0.50 m)
tan φ = (0.5/5) or φ = 5.710 ; 2(2000 N) sin φ = W
W = 4000 sin 5.71; W = 398 N.
*4-12. An 80-N traffic light is supported at the midpoint of a 30-m length of cable between to poles. Find the tension in each cable segment if the cable sags a vertical distance of 1 m.
h = 1 m; Tan φ = (1/15); φ = 3.810
T...
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