Part A If the coordinate direction angles for F3-670 lb are = 115°, = 4s* and =

Question

Part A If the coordinate direction angles for F3-670 lb are = 115°, = 4s* and = 55°, determine the magnitude of the resultant force acting on the eyebolt. (Figure 1) Express your answer to three significant figures and include the appropriate units. h-1 Value Units Value Units Submit My Answers Give Up Not completed before the time limit Determine coordinate direction angle of the resultant force. Express your answer to three significant figures and include the appropriate units. IA Value Units aValue Units Submit My Answers Give Up Not completed before the time limit

Explanation / Answer

Given,

Force, F3 = 670 lb

Angles, = 115° = 45°    = 55°

Let us assume that resultant force be F

Now, we will find the total components of forces along x-direction:

So, Component of F along x-direction = F1cos30°+ F2cos90° + F3cos115° = 700 x 0.866 + 600 x 0 + 670 x (-0.42)

= 324.8 lb

Similarly,  Component of F along y-direction = F1sin30° + F2x(4/5) + F3cos45° = 700 x 0.5 + 600 x (4/5) + 670 x 0.707 = 1303.69 lb

Component of F along Z-direction = F1cos90° + F2x(3/5) + F3cos55° = 700 x 0 + 600 x (3/5) + 670 x 0.57 = 744.3 lb

Hence, Total magnitude of the force, F = sqrt(324.82 + 1303.692 + 744.32) lb = 1535.93 lb

So, = cos-1(Magnitude of x-component of F/ 1535.93) = cos-1(324.8/1535.93) = 77.79°

= cos-1(Magnitude of y-component of F/ 1535.93) = cos-1(1303.69/1535.93) = 31.92°

= cos-1(Magnitude of z-component of F/ 1535.93) = cos-1(744.3/ 1535.93) = 61.01°

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