Explanation Given information: The unstretched length of the cord ( l 0 ) is 1 ft . The weight of the beanbag and holder at C ( W ) is 4 lb . The velocity of the beanbag ( v 2 ) is 31 ft / s . The member OBC is inclined at angle ( α ) is 30 ° . The length of the member OB ( l O B ) is 2 ft . The length of the member BC ( l B C ) is 1ft . The horizontal distance of OA ( x ) is 2 ft . The bag is released at angle ( θ ) is 45 ° . The acceleration due to gravity ( g ) is 32.2 ft / s 2 . Calculation: Show the free body diagram of the given system as in Figure (1). Calculate the mass of the cord ( m ) using the formula: W = m g m = W g Substitute 4 lb for W and 32.2 ft / s 2 for g . m = 4 32.2 = 0.1242 lb ⋅ s 2 / ft Consider the position 1. Calculate the length of AB ( l 1 ) using the law of cosines: l 1 = ( l O B ) 2 + ( l O A ) 2 + ( 2 × l O B × l O A × cos α ) Substitute 2 ft for l O B , 2 ft for l O A , and 30 ° for α . l 1 = ( 2 ) 2 + ( 2 ) 2 + ( 2 × 2 × 2 × cos 30 ° ) = 3.864 ft Calculate the stretch in cord ( s 1 ) using the relation: s 1 = l 1 − l 0 Substitute 3.864 ft for l 1 and 1 ft for l 0 . s 1 = 3.864 − 1 = 2.864 ft Calculate the height of bean bag ( h 1 ) using the relation: h 1 = − ( l O B + l B C ) sin α Substitute 2 ft for l O B , 1ft for l B C and 30 ° for α . h 1 = − ( 2 + 1 ) sin 30 ° = − 1.5 ft Calculate the potential energy stored in the cord due to elongation at position 1 ( V 1 ) e using the formula: ( V 1 ) e = 1 2 k s 1 2 Substitute 2.864 ft for s 1 . ( V 1 ) e = 1 2 × k × ( 2.864 ) 2 = 4.10125 k Calculate the potential energy stored in the cord due to gravitation at position 1 ( V 1 ) g using the formula: ( V 1 ) g = m g h 1 Substitute 0.1242 lb ⋅ s 2 / ft for m , 32.2 ft / s 2 for g and − 1.5 ft for h 1 . ( V 1 ) g = 0.1242 lb ⋅ s 2 / ft × 32.2 × ( − 1.5 ) = − 5.99886 lb ⋅ ft Calculate the potential energy stored in rod in position 1 ( V 1 ) using the relation: V 1 = ( V 1 ) e + ( V 1 ) g Substitute 4.10125 k for ( V 1 ) e and − 5.99886 lb ⋅ ft for ( V 1 ) g . V 1 = 4.10125 k − 5.99886 Here, the kinetic energy in the position 1 ( T 1 ) is zero. Consider the position 2. Here, the length ( l O B ' ) and ( l O B ) are same of 2 ft . Calculate the length of AB’ ( l 2 ) using the law of cosines: l 2 = ( l O B ' ) 2 + ( l O A ) 2 − ( 2 × l O B ' × l O A × cos θ ) Substitute 2 ft for l O B ' , 2 ft for l O A and 45 ° for θ

11th Edition

ISBN: 9781259633133

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 13.2, Problem 13.67P

To determine

Find the spring constant (k).

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13.2, Problem 13.66P

Chapter 13.2, Problem 13.68P

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 13 Solutions

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

Ch. 13.1 - Block A is traveling with a speed v0 on a smooth...Ch. 13.1 - A 400-kg satellite is placed in a circular orbit...Ch. 13.1 - Prob. 13.2P Ch. 13.1 - Prob. 13.3P Ch. 13.1 - A 500-kg communications satellite is in a circular...Ch. 13.1 - Prob. 13.5P Ch. 13.1 - 13.6 In an ore-mixing operation, a bucket full of...Ch. 13.1 - Prob. 13.7P Ch. 13.1 - A 2000-kg automobile starts from rest at point A...Ch. 13.1 - Prob. 13.9P

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.

Find the spring constant ( k ).

Solution Summary: The author shows the free body diagram of the given system as in Figure (1).

Concept explainers

Videos

Find the spring constant (k).

Want to see the full answer?

Chapter 13 Solutions