5.10 . In Fig. E5.10 the weight w is 60.0 N. (a) What is the ten- sion in the diagonal string? (b) Find the magnitudes of the hori- zontal forces F, and Fz that must be applied to hold the system in the position shown. Figure E5.10 90.0 90.0° 45.0 90.0
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- Choose the coefficient of friction wisely. Relevant information is given in the textbook. a. What is the maximum frictional force in the knee joint of a person who supports 70.0 kg of her mass on that knee? Fmax 686.7 XN b. During strenuous exercise it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? (The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain.) Fmax = NH5.(a) Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 7.00 x 105 kg hangs motionless on it. (Enter a number.) N (b) Calculate the tension (in N) in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in the figure. 5.0° 5,0° T. TR T I Ta w The strand sags at an angle of 12.0° below the horizontal. (Enter a number.) N Compare this with the tension in the vertical strand (find their ratio). (Enter a number.) (tension in horizontal strand)/(tension in vertical strand) =
- Problem 4: A ball of mass m is connected to two rubber bands of length L, cach under temsion T, as shown in the Figure. Assume the tension does not change. (a) The ball of mass m is in equilibrium. Draw a free-body diagram on m. (b) Since the ball of mass m is in equilibrium, Enet y = 0. Use Fnet.y=0 to find the equilibrium position for the ball yo. Hint: sin 0=yo/L, Answer: yo = mgL/(2T'). |Yo L. T. L т (c) The ball of mass m is now displaced slightly from equilibrium, as shown in the figure below. Draw a free-body diagram on m. (d) Use Newton's 2nd Law: Fnet,y = ma = -m to show that the motion of m satisfies the SHM differential equation + w²u = 0, where u = y – Yo and w = since the downward gravitational force on the ball is larger than the two upward tension forces on the ball. Hint: sin 0 = y/L. Also, from part (b), yo = mgL/(2T). We can rewrite this expression as mg = 2T(yo/L). Use this expression to substitute for mg in Newton's 2nd Law: Fnet.y = -ma = -my | 2T. For this problem…3. The force F required to move a box by pulling on an attached rope that forms an angle w with the horizontal is given by fmg F(w) = cOSw + f sin w where f, m, g > 0 are all constant. (a) Show that the angle wo that minimizes the required force to set the box in motion is given by arctan(f) (b) Show that this minimal force F(wo) is proportional to (a constant multiple of)Refer to the question.
- A horizontal force with magnitude of 52.0N on a box and produces acceleration with magnitude 3.60 m/s^2. What is the mass of the box?1F=10k 30 5=12k 5=18k Calculate the Fx and Fy components of all the the forces and determine the resultant force. 券A 1.5 kg block resting ina rough flat surface is pulled by a string. The string makes an angle of 20 degress with the horizontal. The coefficient of static friction between the blodk and the surface is 0.35. What is the maximum tension in the string to keep the block unmoved?