Problem 3 The beam has a rectangular cross section as shown. Determine the largest intensity w of the uniform distributed load so that the bending stress in the beam does not exceed max=10 MPa. W 50 mm -2 m- -2 m -2 m [150 mm
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- A copper rod with length 1.4 m and cross-sectional area 2.0 cm2 is fastened to a steel rod of length L and cross-sectional area 1.0 cm2. The compound structure is pulled on each side by two forces of equal magnitude 6.00 104 N (Fig. P14.57). Find the length L of the steel rod if the elongations (L) of the two rods are equal. Use the values Ysteel = 2.0 1011 Pa and YCu = 1.1 1011 Pa. FIGURE P14.57The temperature of the atmosphere oscillates from a maximum near noontime and a minimum near sunrise. Would you consider the atmosphere to be in stable or unstable equilibrium?In the short story The Pit and the Pendulum by 19th-century American horror writer Edgar Allen Poe, a man is tied to a table directly below a swinging pendulum that is slowly lowered toward him. The bob of the pendulum is a 1-ft steel scythe connected to a 30-ft brass rod. When the man first sees the pendulum, the pivot is roughly 1 ft above the scythe so that a 29-ft length of the brass rod oscillates above the pivot (Fig. P16.39A). The man escapes when the pivot is near the end of the brass rod (Fig. P16.39B). a. Model the pendulum as a particle of mass ms 5 2 kg attached to a rod of mass mr 5 160 kg. Find the pendulums center of mass and rotational inertia around an axis through its center of mass. (Check your answers by finding the center of mass and rotational inertia of just the brass rod.) b. What is the initial period of the pendulum? c. The man saves himself by smearing food on his ropes so that rats chew through them. He does so when he has no more than 12 cycles before the pendulum will make contact with him. How much time does it take the rats to chew through the ropes? FIGURE P16.39
- (a) A hanging spring stretches by 35.0 cm when an object of mass 450 g is hung on it at rest. In this situation, we define its position as x = 0. The object is pulled down an additional 18.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later? (b) Find the distance traveled by the vibrating object in part (a), (c) What If? Another hanging spring stretches by 35.5 cm when an object of mass 440 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 18.0 cm and released from rest to oscillate without friction. Find its position 84.4 s later, (d) Find the distance traveled by the object in part (c). (e) Why are the answers to parts (a) and (c) so different when the initial data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close? Does this circumstance reveal a fundamental difficulty in calculating the future?Figure P13.74 shows a crude model of an insect wing. The mass m represents the entire mass of the wing, which pivots about the fulcrum F. The spring represents the surrounding connective tissue. Motion of the wing corresponds to vibration of the spring. Suppose the mass of the wing is 0.30 g and the effective spring constant of the tissue is 4.7 104 N/m. If the mass m moves up and down a distance of 2.0 mm from its position of equilibrium, what is the maximum speed of the outer tip of the wing? Figure P13.74An object of mass m1 = 9.00 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P12.67a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m (see Fig. P12.67b). The system is then released, and both objects start moving to the right on the frictionless surface. (a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P12.67c) and moves to the right with speed v. Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in Fig. P12.67d)? Figure P12.67
- Some people modify cars to be much closer to the ground than when manufactured. Should they install stiffer springs? Explain your answer.Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction s=0.100 . (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is k=0.0850 , what total distance does it travel before stopping? Assume it starts at the maximum amplitude.(a) If frequency is not constant for some oscillation, can the oscillation be SHM? (b) Can you think of any examples of harmonic motion where the frequency may depend on the amplitude?
- The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?A pendulum of length L and mass M has a spring of force constant k connected to it at a distance h below its point of suspension (Fig. P12.65). Find the frequency of vibration of the system for small values of the amplitude (small ). Assume the vertical suspension rod of length L is rigid, but ignore its mass. Figure P12.65A lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77