The resultant force of two forces is a vector sum of the vectors of these forces. A vector sum can be found by the law of cosines:
F r e s u l t a n t 2 = F 1 2 + F 2 2 + 2 F 1 F 2 c o s ( α − β ) F_{resultant}^2=F_1^2+F_2^2+2F_1F_2cos(\alpha-\beta) F res u lt an t 2 = F 1 2 + F 2 2 + 2 F 1 F 2 cos ( α − β )
After setting given values we have:
F r e s u l t a n t = 1 0 4 + 1 0 4 + 2 ∗ 1 0 2 ∗ 1 0 2 ∗ ( − 1 / 2 ) F_{resultant}=\sqrt{10^4+10^4+2*10^2*10^2*(-1/2)} F res u lt an t = 1 0 4 + 1 0 4 + 2 ∗ 1 0 2 ∗ 1 0 2 ∗ ( − 1/2 )
F r e s u l t a n t = 1 0 4 + 1 0 4 − 1 0 4 F_{resultant}=\sqrt{10^4+10^4-10^4} F res u lt an t = 1 0 4 + 1 0 4 − 1 0 4
F r e s u l t a n t = 1 0 4 F_{resultant}=\sqrt{10^4} F res u lt an t = 1 0 4
F r e s u l t a n t = 100 N F_{resultant}=100N F res u lt an t = 100 N
The action angle of the resultant force equals:
γ = ( α + β ) / 2 = ( 170 ° + 50 ° ) / 2 = 110 ° \gamma=(\alpha+\beta)/2=(170°+50°)/2=110° γ = ( α + β ) /2 = ( 170° + 50° ) /2 = 110°
Answer: The resultant force equals 100N at 110°.
Comments