Question

Four concurrent forces1kN, 2kN, 3kN and 4kN acting at an angle of 20°, 63°, 95°,150° from positive x axis. Determine their resultant in kN.​

Answer 1

Answer:

The resultant of four forces, which are acting at a point, is along Y-axis. The magnitudes of forces F1, F3, F4 are 10KN, 20KN and 40KN respectively. The angle made by 10KN, 20KN and 40KN with X-axis are 300, 900 and 1200 respectively. Find the magnitude and direction of force F2, if resultant is 72KN. Read more on Sarthaks.com - https://www.sarthaks.com/507010/resultant-forces-which-acting-point-along-magnitudes-forces-10kn-20kn-40kn-respectively

Answer 2

Concept Introduction:-

It could take the shape of a word or a number representation of the quantity's arithmetic value.

Given Information:-

We have been given that Four concurrent forces $1$ kN, $2$ kN, $3$ kN and $4$ kN acting at an angle of $20\°$, $63\°$, $95\°$, $150\°$ from positive $x$ axis.

To Find:-

We have to find that their resultant in kN.​

Solution:-

According to the problem

Horizontal forces

$\begin{aligned}&=10 \cos 20^{\circ}+20 \cos 63^{\circ}+30 \cos 95^{\circ}+40 \cos 150^{\circ} \\&=9.397+9.08-2.615-34.641 \\&=-18.779 \mathrm{kN}\end{aligned}$

Vertical forces

$\begin{aligned}&=10 \sin 20^{\circ}+20 \sin 63^{\circ}+30 \sin 95^{\circ}+40 \sin 150^{\circ} \\&=3.42+17.82+29.886+20 \\&=71.126 \mathrm{kN}\end{aligned}$

$\begin{aligned}&\text { resultant in } \mathrm{kN} .=\sqrt{(-18.779)^{2}+(71.126)^{2} }\\&=73.563 \mathrm{kN}\end{aligned}$

their resultant in $\mathrm{kN} .=73.563$

Final Answer:-

The correct answer is their resultant in kN.​ is $73.563$.

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