Problem 10-4. The drawing shows a 160-kg crate hanging from the end of a steel bar. The length of the bar is 0.10 m, and its cross-sectional area is 3.2 × 10-4 m2. Neglect the weight of the bar itself and determine (a) the shear stress on the bar and (b) the vertical deflection y of the right end.
(a) The force on the end of the bar is the weight of the crate. This is mg = 160 kg × 9.8 m/s2 = 1570 N. The stress is defined as the force divided by the area over which it is applied, so it is 1570 N / 3.2 × 10-4 m2 = 4.9 × 106 N/m2.
(b) The vertical deflection is found by solving the elastic equation
Problem 10-8. A copper cylinder and a brass cylinder are stacked end to end. Each cylinder has a radius of 0.25 cm. A compressive force of F = 6500 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.
You have to realize that the compressive force is passed down the stack, and is experienced by both parts, just as the tension in a spring is the same at all points along its length. Calculate the decrease in length of each cylinder separately, and add them together.
For the copper cylinder
The same formula for the brass cylinder gives a decrease in length of
The total change in length is then 2.7 × 10-4 m.
Problem 10-28. A spring (k = 830 N/m) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.60 m/s2?
Write down Newton's second law for the accelerating object (taking upwards as positive)
Problem 10-43. Suppose that an object on a vertical spring oscillates up and down at a frequency of 5.00 Hz. By how much would this object, hanging at rest, stretch the spring?
From the value of the frequency, you can get some information about the mass of the object and the spring constant of the spring, but there is not enough information to get the actual values of them.
The extension of the spring when the object hangs at rest is found from the equilibrium condition
Problem 10-50. The spring constant for a spring in a dart gun is 1400 N/m. When the gun is cocked, the spring is compressed by 0.075 m. What is the speed of a 2.4 × 10-2-kg dart when it leaves the gun horizontally?
The elastic potential energy stored in the compressed spring is transformed into the kinetic energy of the dart. So