Explanation Given information: The system is in equilibrium when θ 1 = θ 2 = 0 . Calculation: Draw the free-body diagram of the arrangement as in Figure (1). Consider the movement of spring at left end is from a to b , and the right end is from a ′ to b ′ . Find the elongation of the spring ( s ) using the relation. s = a ′ b ′ − a b = r θ 1 − r θ 2 = r ( θ 1 − θ 2 ) Find the potential energy ( V ) using the relation. V = V g + V s = 1 2 k s 2 + P l cos θ 1 − W l cos θ 2 = 1 2 k r 2 ( θ 1 − θ 2 ) 2 + P l cos θ 1 − W l cos θ 2 (1) Here, the spring constant is k . Differentiate the Equation (1) with respect to θ 1 . ∂ V ∂ θ 1 = 2 × 1 2 k r 2 ( θ 1 − θ 2 ) − P l sin θ 1 − 0 = k r 2 ( θ 1 − θ 2 ) − P l sin θ 1 (2) Differentiate the Equation (2) with respect to θ 1 . ∂ 2 V ∂ θ 1 2 = k r 2 − P l cos θ 1 Differentiate the equation (2) with θ 2 to find the derivative of ∂ 2 V ∂ θ 1 ∂ θ 2 . ∂ 2 V ∂ θ 1 ∂ θ 2 = − k r 2 Differentiate the Equation (1) with respect to θ 2 . ∂ V ∂ θ 2 = − 2 × 1 2 k r 2 ( θ 1 − θ 2 ) + 0 − W l sin θ 2 = − k r 2 ( θ 1 − θ 2 ) − W l sin θ 2 (3) Differentiate the Equation (3) with respect to ∂ θ 2 . ∂ 2 V ∂ θ 2 2 = k r 2 + W l cos θ 2 Condition 1: When the equilibrium is stable, θ 1 = θ 2 = 0 . Substitute 0 for θ 1 and 0 for θ 2 in Equation (2). ∂ V ∂ θ 1 = k r 2 ( 0 − 0 ) − P l sin 0 = 0 Substitute 0 for θ 1 and 0 for θ 2 in Equation (3). ∂ V ∂ θ 2 = − k r 2 ( 0 − 0 ) − W l sin 0 = 0 ∂ V ∂ θ 1 = ∂ V ∂ θ 2 = 0 The condition is satisfied

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Gen…

Videos

Question

Chapter 10.2, Problem 10.99P

To determine

The range of values of P for which the equilibrium is stable.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.2, Problem 10.98P

Chapter 10.2, Problem 10.100P

Students have asked these similar questions

Q Derive (continuity equation)? I want to derive clear mathematics.

motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

First monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³

Chapter 10 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

Ch. 10.1 - Determine the vertical force P that must be...Ch. 10.1 - Determine the horizontal force P that must be...Ch. 10.1 - Prob. 10.3P Ch. 10.1 - Prob. 10.4P Ch. 10.1 - Prob. 10.5P Ch. 10.1 - A spring of constant 15 kN/m connects points C and...Ch. 10.1 - The two-bar linkage shown is supported by a pin...Ch. 10.1 - Prob. 10.8P Ch. 10.1 - Prob. 10.9P Ch. 10.1 - Prob. 10.10P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.