PROBLEMAS DE FISICA

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 represent each 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 cord supports 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. If the 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 600

B = (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 for Fig. 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 block from 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 weight of the sign? (h = 0.50 m)

tan  = (0.5/5) or  = 5.710 ; 2(2000 N) sin  = 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  = (1/15);  = 3.810

T sin  + T sin  = 80 N; 2T sin  = 80 N

Solution to 4-12 (Cont.): ; T = 601 N

*4-13. The ends of three 8-ft studs are nailed together forming a tripod with an apex that is 6ft above the ground. What is the compression in each of these studs if a 100-lb weight is hung from the apex?

Three upward components Fy hold up the 100 lb weight:

3 Fy = 100 lb; Fy = 33.3 lb sin  = (6/8);  = 48.90

F sin 48.90 = 33.3 lb; F = 44.4 lb, compression

*4-14. A 20-N picture is hung from a nail as in Fig. 4-20, so that the supporting cords make an angle of 600. What is the tension of each cord segment?

According to Newton’s third law, the force of frame on nail (20 N)

is the same as the force of the nail on the rope (20 N , up).

Fy = 0; 20 N = Ty + Ty; 2Ty = 20 N; Ty = 10 N

Ty = T sin 600; So T sin 600 = 10 N, and T = 11.5 N.

Friction

4-15. A horizontal force of 40 N will just start an empty 600-N sled moving across packed snow. After motion is begun, only 10 N is needed to keep motion at constant speed. Find the coefficients of static and kinetic friction.

s = 0.0667; k = 0.016

4-16. Suppose 200-N of supplies are added the sled in Problem 4-13. What new force is needed to drag the sled at constant speed?

N= 200 N + 600 N = 800 N; Fk = kN = (0.0167)(800 N); Fk = 13.3 N

4-17. Assume surfaces where s = 0.7 and k = 0.4. What horizontal force is needed to just start a 50-N block moving along a wooden floor. What force will move it at constant speed?

Fs = s¬N = (0.7)(50 N) = 35 N ; Fk = s¬N = (0.4)(50 N) = 20 N

4-18. A dockworker finds that a horizontal force of 60 lb is needed to drag a 150-lb crate across the deck at constant speed. What is the coefficient of kinetic friction?

; k = 0.400

4-19. The dockworker in Problem 4-16 finds that a smaller crate of similar material can be dragged at constant speed with a horizontal force of only 40 lb. What is the weight of this crate?

Fk = s¬N = (0.4)W = 40 lb; W = (40 lb/0.4) = 100 lb; W = 100 lb.

4-20. A steel block weighing 240 N rests on level steel beam. What horizontal force will move the block at constant speed if the coefficient of kinetic friction is 0.12?

Fk = s¬N = (0.12)(240 N) ; Fk = 28.8 N.

4-21. A 60-N toolbox is dragged horizontally at constant speed by a rope making an angle of 350 with the floor. The tension in the rope is 40 N. Determine the magnitude of the friction force and the normal force.

Fx = T cos 350 – Fk = 0; Fk = (40 N) cos 350 = 32.8 N

Fy = N + Ty – W = 0; N = W – Ty = 60 N – T sin 350

N = 60 N – (40 N) sin 350; N = 37.1 N Fk = 32.8 N

4-22. What is the coefficient of kinetic friction for the example in Problem 4-19?

k = 0.884

4-23. The coefficient of static friction for wood on wood is 0.7. What is the maximum angle for an inclined wooden plane if a wooden block is to remain at rest on the plane?

Maximum angle occurs when tan  = s; s = tan  = 0.7;  = 35.00

4-24. A roof is sloped at an angle of 400. What is the maximum coefficient of static friction between the sole of the shoe and the roof to prevent slipping?

Tan  = k; k = Tan 400 =0.839; k = 0.839

*4-25. A 200 N sled is pushed along a horizontal surface at constant speed with a 50-N force that makes an angle of 280 below the horizontal. What is the coefficient of kinetic friction?

Fx = T cos 280 – Fk = 0; Fk = (50 N) cos 280 = 44.1 N

Fy = N - Ty – W = 0; N = W + Ty = 200 N + T sin 280

N = 200 N + (50 N) sin 350; N = 223 N

k = 0.198

*4-26. What is the normal force on the block in Fig. 4-21? What is the component of the weight acting down the plane?

Fy = N - W cos 430 = 0; N = (60N) cod 430 = 43.9 N

Wx = (60 N) sin 350; Wx = 40.9 N

*4-27. What push P directed up the plane will cause the block in Fig. 4-21 to move up the plane with constant speed? [From Problem 4-23: N = 43.9 N and Wx = 40.9 N]

Fk = kN = (0.3)(43.9 N); Fk = 13.2 N down plane.

Fx = P - Fk – Wx = 0; P = Fk + Wx; P = 13.2 N + 40.9 N; P = 54.1 N

*4-28. If the block in Fig. 4-21 is released, it will overcome static friction and slide rapidly down the plane. What push P directed up the incline will retard the downward motion until the block moves at constant speed? (Note that F is up the plane now.)

Magnitudes of F , Wx, and N are same as Prob. 4-25.

Fx = P +Fk – Wx = 0; P = Wx - Fk; P = 40.9 N - 13.2 N

P = 27.7 N directed UP the inclined plane

Challlenge Problems

*4-29. Determine the tension in rope A and the compression B in the strut for Fig. 4-22.

Fy = 0; By – 400 N = 0;

Fx = 0; Bx – A = 0; A = B cos 600

A = (462 N) cos

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