Discussion :: Force Vectors - General Questions (Q.No.16)
|Monica Ann said: (Jun 18, 2011)|
|First, solve for the angle of F2.
Let's have the angle as angle α (alpha).
That could be solved by using the sine law.
Equating this gives us, α=22.61986495 degrees
Now, we can solve for F1x, F1y, F2x, F2y, Rx, Ry, R (resultant) and θ.
F1x= 259.8076211 N to the right
F1y= 150 N, up
F2x= 240 N to the left
F2y= 100 N, up
= [259.8076211 + (-240)]N
*Note that we have -F2x since the force is to the left.
Rx= 19.80762114 N
Ry= F1y + F2y
= [150 + 100]N
Ry= 250 N
R=*squareroot of [Rx*squared + Ry*squared]
=*squareroot of [(19.80...)*squared + (250)*squared]
=*squareroot of (392.3418552 + 62500)
=*squareroot of (62892.34186)
R=250.7834561 N, which could be rounded to 251 N.
θ= arctan (Ry/Rx)
= arctan (250/19.80...)
= arctan (12.62140457)
θ= 85.46989102 degrees, which could be rounded to 85.5 degrees.
Answer: 251 N, θ=85.5 degrees.
|Khuliso said: (Sep 12, 2011)|
|The calculation are good, but is the any way we can do this?|
|Rye said: (Jan 10, 2013)|
|Solving for the component of x
fx = [300cos30]+[-260x12/13]
fx = 20 N
Solving for fy:
Therefore R=20squared+250square then taking its squareroot
Solving for angle:
Angle =arctan fy/fx
|Yoseph said: (Dec 27, 2013)|
|Clear but if it will be in parallelogram.|
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