The bracket is subjected to the two forces shown. Express each force in Cartesia
ID: 1717469 • Letter: T
Question
The bracket is subjected to the two forces shown. Express each force in Cartesian vector form and then determine the resultant force F_R. Given that F_2 = 500 N Find the magnitude of the resultant force. Express your answer to three significant figures and Include the appropriate units. Find coordinate director angle alpha of the resultant force. Express your answer to three significant figures and Include the appropriate unite. Find coordinate direction angle beta of the resultant force. Express your answer to three significant figures and include the appropriate units.Explanation / Answer
>> Now, Considering Force F1,
Magnitude F1 = 250 N
and angle with y-axis = 25 degree
angle with x - axis = (90 - 35)= 55 degree
and, angle with z-axis = 90 degree
=> F1 = 250(cos55 i + cos25 j + cos90 k)
=> F1 = 143.39 i + 226.58 j (N) .......(1)....
>> Now, Considering Force F2,
Magnitude F1 = 500 N
and angle with y-axis = 45 degree
angle with x - axis = 120 degree
and, angle with z-axis = 60 degree
=> F2 = 500(cos120 i + cos45 j + cos60 k)
=> F2 = - 250 i + 353.55 j + 250 k (N) .......(2)....
>> Now, as Fr = Resulatnt Force = F1 + F2
=> Fr = 143.39 i + 226.58 j - 250 i + 353.55 j + 250 k
=> Fr = - 106.61 i + 580.13 j + 250 k (N)......
Magnitude of resultant force, |Fr| = [ -106.612 + 580.132 + 2502 ]1/2
= 640.637 N .............ANSWER........
>> As, co-ordinate direction angle = angle made by resultant vector with x-axis
>> As, Fr = - 106.61 i + 580.13 j + 250 k and magnitude = 640.637 N
=> Unit Vector along Fr = (- 106.61 i + 580.13 j + 250 k)/640.637
=> Unit Vector along Fr = - 0.166 i + 0.91 j + 0.39 k
>> So, directon cosines are < l, m , n > = < cos, cos, cos > = < -0.166, 0.91, 0.39 >
=> cos = -0.166
=> = 99.555 degree = co-ordinate direction angle ........ANSWER..........
>> Also, cos= 0.91
=> = 24.495 degree = co-ordinate direction angle ........ANSWER..........
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