Question

A hook is subjected to three forces as shown in the figure below. F3 = 105 lb F2 = 70 lb 60° 45° F1 = 60 lb The force vector F3 in lb is represented as The magnitude of the resultant force in lb is -105 cos(75) i + 105 sin(75) j 105 cos(75) i + 105 sin(75) j -105 sin(75) i + 105 cos(75) j 105 sin(75) i + 105 cos(75) j 171 143 196 83

          A hook is subjected to three forces as shown in the figure below.
F3 = 105 lb
F2 = 70 lb
60°
45°
F1 = 60 lb
The force vector F3 in lb is represented as
The magnitude of the resultant force in lb is
-105 cos(75) i + 105 sin(75) j  105 cos(75) i + 105 sin(75) j  -105 sin(75) i + 105 cos(75) j  105 sin(75) i + 105 cos(75) j
171  143  196  83

A hook is subjected to three forces as shown in the figure below.
F3 = 105 lb
F2 = 70 lb
60°
45°
F1 = 60 lb
The force vector F3 in lb is represented as
The magnitude of the resultant force in lb is
-105 cos(75) i + 105 sin(75) j 105 cos(75) i + 105 sin(75) j -105 sin(75) i + 105 cos(75) j 105 sin(75) i + 105 cos(75) j
171 143 196 83

Added by Deanna A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Monisha Shukla

Step 1

F = 105 lb can be broken down into: Fx = 105 lb * cos(75°) Fy = 105 lb * sin(75°) F2 = 701 lb can be broken down into: F2x = 701 lb * cos(0°) F2y = 701 lb * sin(0°) S F = 60 lb can be broken down into: S Fx = 60 lb * cos(180°) S Fy = 60 lb * sin(180°) Show more…

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A hook is subjected to three forces as shown in the figure below. F = 105 lb F2 = 701 lb S F = 60 lb The force vector F in lb is represented as The magnitude of the resultant force in lb is 105cos(75i) + 105sin(75j) - 105sin(75i) + 105cos(75j) + 105sin(75i) + 105cos(75j) + 105cos(75i) + 105sin(75j) 96 83