Explanation Given: First of all, a C = a n = v C 2 ρ Where, ρ = 3 ft ← + ∑ F x = m C a C : T C A sin 40 ∘ + T C B sin 15 ∘ = W C g v C 2 ρ … (1) + ↑ ∑ F y = 0 : T C A cos 40 ∘ + T C B cos 15 ∘ − W C = 0 … (2) Eq. (2) implies, When T C B = ( T C B ) max , T C A = ( T C A ) max When T C B = ( T C B ) min , T C A = ( T C A ) min Case 1: T C A is maximum Let T C A = 26 lb Eq. (2) ( 26 lb ) cos 40 ∘ − T C B cos 15 ∘ − ( 12 lb ) = 0 Or: T C B = 8.1964 lb ( T C B ) ( T C A ) max < 26 lb (OK) [ ( T C B ) max = 8.1964 lb ] Eq. (1) ( v C 2 ) ( T C A ) max = ( 32.2 ft / s 2 ) ( 3 ft ) 12 lb ( 26 sin 40 ∘ + 8.1964 sin 15 ∘ ) lb Or: ( v C ) ( T C A ) max = 12.31 ft / s Now we form ( cos 15 ∘ ) ( Eq

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 12.1, Problem 12.42P

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Range of allowable values of v if both wires are to remain taut and if tension in either of the wires is not to exceed 26 lb.

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Chapter 12.1, Problem 12.41P

Chapter 12.1, Problem 12.43P

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Vector Mechanics for Engineers: Dynamics

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Range of allowable values of v if both wires are to remain taut and if tension in either of the wires is not to exceed 26 lb .

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Range of allowable values of v if both wires are to remain taut and if tension in either of the wires is not to exceed 26 lb.

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