Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
11th Edition
ISBN: 9780073398242
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 10.2, Problem 10.100P

(a)

To determine

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

(a)

Expert Solution
Check Mark

Answer to Problem 10.100P

The range of values of P for which the equilibrium position is stable is P<10lb_.

Explanation of Solution

Given information:

The system is in equilibrium when θ1=θ2=0.

The value of spring constant is k=20lb/in..

The radius of the drums is r=3in..

The length of the rods AB and CD is l=6in..

The weight acting at point A is W=15lb.

Calculation:

Draw the free-body diagram of the arrangement as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 10.2, Problem 10.100P

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=abab=rθ1rθ2=r(θ1θ2)

Find the potential energy (V) using the relation.

V=Vg+Vs=12ks2+Plcosθ1Wlcosθ2=12kr2(θ1θ2)2+Plcosθ1Wlcosθ2 (1)

Here, the spring constant is k.

Differentiate the Equation (1) with respect to θ1.

Vθ1=2×12kr2(θ1θ2)Plsinθ10=kr2(θ1θ2)Plsinθ1 (2)

Differentiate the Equation (2) with respect to θ1.

2Vθ12=kr2Plcosθ1

Differentiate the equation (2) with θ2 to find the derivative of 2Vθ1θ2.

2Vθ1θ2=kr2

Differentiate the Equation (1) with respect to θ2.

Vθ2=2×12kr2(θ1θ2)+0Wlsinθ2=kr2(θ1θ2)Wlsinθ2 (3)

Differentiate the Equation (3) with respect to θ2.

2Vθ22=kr2+Wlcosθ2

Condition 1:

When the equilibrium is stable, θ1=θ2=0.

Substitute 0 for θ1 and 0 for θ2 in Equation (2).

Vθ1=kr2(00)Plsin0=0

Substitute 0 for θ1 and 0 for θ2 in Equation (3).

Vθ2=kr2(00)Wlsin0=0

Vθ1=Vθ2=0

The condition is satisfied. The equilibrium is stable.

Condition 2:

Check the condition,

(2Vθ1θ2)22Vθ122Vθ22<0

Substitute kr2 for 2Vθ1θ2, (kr2Plcosθ1) for 2Vθ12, and (kr2+Wlcosθ2) for 2Vθ22.

(kr2)2(kr2Plcosθ1)(kr2+Wlcosθ2)<0

Substitute 0 for θ1 and 0 for θ2.

(kr2)2(kr2Plcos0°)(kr2+Wlcos0°)<0(kr2)2(kr2Pl)(kr2+Wl)<0k2r4k2r4kr2Wl+kr2Pl+PWl2<0kr2W+P(kr2+Wl)<0

P<kr2W(kr2+Wl)P<kr2l(W(kr2l+W)) (4)

Condition 3:

2Vθ12>0kr2Plcosθ1>0

Substitute 0 for θ1.

kr2Plcos0°>0P<kr2l

Refer to all the conditions,

The minimum value of P is 0.

The maximum value of P is Pmax<kr2l(W(kr2l+W)).

Substitute 20lb/in. for k, 6 in. for l, 15 lb for W, and 3 in. for r in Equation (4).

P<20×326(15(20×326+15))<20×96(1545)<10lb

Thus, the range of values of P for which the equilibrium position is stable is P<10lb_.

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 10.100P

The range of values of P for which the equilibrium position is stable is P<20lb_.

Explanation of Solution

Given information:

The system is in equilibrium when θ1=θ2=0.

The value of spring constant is k=20lb/in..

The radius of the drums is r=3in..

The length of the rods AB and CD is l=6in..

The weight acting at point A is W=60lb.

Calculation:

Substitute 20lb/in. for k, 6 in. for l, 60 lb for W, and 3 in. for r in Equation (4).

P<20×326(60(20×326+60))<20×96(6090)<20lb

Thus, the range of values of P for which the equilibrium position is stable is P<20lb_.

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Chapter 10 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 10.1 - Prob. 10.11PCh. 10.1 - Prob. 10.12PCh. 10.1 - Prob. 10.13PCh. 10.1 - Prob. 10.14PCh. 10.1 - Prob. 10.15PCh. 10.1 - Prob. 10.16PCh. 10.1 - Prob. 10.17PCh. 10.1 - Prob. 10.18PCh. 10.1 - Prob. 10.19PCh. 10.1 - Prob. 10.20PCh. 10.1 - Prob. 10.21PCh. 10.1 - A couple M with a magnitude of 100 Nm isapplied as...Ch. 10.1 - Rod AB is attached to a block at A that can...Ch. 10.1 - Solve Prob. 10.23, assuming that the 800-N force...Ch. 10.1 - Prob. 10.25PCh. 10.1 - Prob. 10.26PCh. 10.1 - Prob. 10.27PCh. 10.1 - Prob. 10.28PCh. 10.1 - Prob. 10.29PCh. 10.1 - Two rods AC and CE are connected by a pin at Cand...Ch. 10.1 - Solve Prob. 10.30 assuming that force P is movedto...Ch. 10.1 - Prob. 10.32PCh. 10.1 - Prob. 10.33PCh. 10.1 - Prob. 10.34PCh. 10.1 - Prob. 10.35PCh. 10.1 - Prob. 10.36PCh. 10.1 - Prob. 10.37PCh. 10.1 - Prob. 10.38PCh. 10.1 - Prob. 10.39PCh. 10.1 - Prob. 10.40PCh. 10.1 - Prob. 10.41PCh. 10.1 - The position of boom ABC is controlled by...Ch. 10.1 - Prob. 10.43PCh. 10.1 - Prob. 10.44PCh. 10.1 - Prob. 10.45PCh. 10.1 - Prob. 10.46PCh. 10.1 - Denoting the coefficient of static friction...Ch. 10.1 - Prob. 10.48PCh. 10.1 - Prob. 10.49PCh. 10.1 - Prob. 10.50PCh. 10.1 - Prob. 10.51PCh. 10.1 - Prob. 10.52PCh. 10.1 - Prob. 10.53PCh. 10.1 - Prob. 10.54PCh. 10.1 - Prob. 10.55PCh. 10.1 - Prob. 10.56PCh. 10.1 - Prob. 10.57PCh. 10.1 - Prob. 10.58PCh. 10.2 - Using the method of Sec. 10.2C, solve Prob. 10.29....Ch. 10.2 - Prob. 10.60PCh. 10.2 - Prob. 10.61PCh. 10.2 - Prob. 10.62PCh. 10.2 - Prob. 10.63PCh. 10.2 - Prob. 10.64PCh. 10.2 - Prob. 10.65PCh. 10.2 - Using the method of Sec. 10.2C, solve Prob. 10.38....Ch. 10.2 - Prob. 10.67PCh. 10.2 - Prob. 10.68PCh. 10.2 - Prob. 10.69PCh. 10.2 - Prob. 10.70PCh. 10.2 - Prob. 10.71PCh. 10.2 - Prob. 10.72PCh. 10.2 - Prob. 10.73PCh. 10.2 - Prob. 10.74PCh. 10.2 - Prob. 10.75PCh. 10.2 - Prob. 10.76PCh. 10.2 - Prob. 10.77PCh. 10.2 - Prob. 10.78PCh. 10.2 - Prob. 10.79PCh. 10.2 - Prob. 10.80PCh. 10.2 - Prob. 10.81PCh. 10.2 - A spring AB of constant k is attached to two...Ch. 10.2 - Prob. 10.83PCh. 10.2 - Prob. 10.84PCh. 10.2 - Prob. 10.85PCh. 10.2 - Prob. 10.86PCh. 10.2 - Prob. 10.87PCh. 10.2 - Prob. 10.88PCh. 10.2 - Prob. 10.89PCh. 10.2 - Prob. 10.90PCh. 10.2 - Prob. 10.91PCh. 10.2 - Prob. 10.92PCh. 10.2 - Prob. 10.93PCh. 10.2 - Prob. 10.94PCh. 10.2 - Prob. 10.95PCh. 10.2 - Prob. 10.96PCh. 10.2 - Bars AB and BC, each with a length l and of...Ch. 10.2 - Prob. 10.98PCh. 10.2 - Prob. 10.99PCh. 10.2 - Prob. 10.100PCh. 10 - Determine the vertical force P that must be...Ch. 10 - Determine the couple M that must be applied...Ch. 10 - Prob. 10.103RPCh. 10 - Prob. 10.104RPCh. 10 - Prob. 10.105RPCh. 10 - Prob. 10.106RPCh. 10 - Prob. 10.107RPCh. 10 - Prob. 10.108RPCh. 10 - Prob. 10.109RPCh. 10 - Prob. 10.110RPCh. 10 - Prob. 10.111RPCh. 10 - Prob. 10.112RP
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