Estatica

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COSMOS: Complete Online Solutions Manual Organization System

Chapter 2, Problem 5.

The 200-N force is to be resolved into components along lines a-a′ and
b-b′. (a)Determine the angle α using trigonometry knowing that the
component along a-a′ is to be 150 N. (b) What is the corresponding
value of the component along b-b′ ?SOLUTION:

Using the triangle rule and the Law of Sines

(a)

sin β
sin 45°
=
150 N 200 N

sin β = 0.53033

β = 32.028°

α + β + 45° = 180°

α = 103.0° ⊳(b)

Using the Law of Sines

Fbb′
200 N
=
sin α
sin 45°

Fbb′ = 276 N ⊳

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E.Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.

COSMOS: Complete Online SolutionsManual Organization System

Chapter 2, Problem 12.

To steady a sign as it is being lowered, two cables are attached to the sign
at A. Using trigonometry and knowing thatthe magnitude of P is 70 lb,
determine (a) the required angle α if the resultant R of the two forces
applied at A is to be vertical, (b) the corresponding magnitude ofR.

SOLUTION:

Using the triangle rule and the Law of Sines

(a) Have:

80 lb
70 lb
=
sin α
sin 35°

sin α = 0.65552

α = 40.959°

or α = 41.0° ⊳

(b )β = 180 − ( 35° + 40.959° )

= 104.041°

Then:

R

sin 104.041°

70 lb
sin 35°

=

or R = 118.4 lb⊳

Vector Mechanics for Engineers: Statics andDynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
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