Determine the magnitude and direction of the resultant of the following set of forces acting on a body (i) 200 N inclined 30° with east towards north,(ii) 250 N towards north, (iii) 300 N towards north west(iv) 350 N inclined at 40° with west toward south.
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Answer:
F=86.37 N
Explanation:
Because system of force is in equilibrium so sum of horizontal (ΣFH) and vertical
(ΣFV) component of all forces will be zero.
ΣFH=0
Fcos60° +200cos45° =300+Pcos30°
F=317.16+1.7320P
ΣFV=0
Fsin60° =200sin45° +Psin30°
(317.16+1.7320P) sin60° =200sin45° +Psin30°
274.67+1.5P=141.42+0.5P
P=−133.25 N
F=317.16+1.7320(−133.25)
F=86.37 N
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