Three forces of magnitude 6 N, 5 N, and 10 N are acting on the same load. The directions of the forces are 70°, 160°, and 240° respectively. Find the resultant force.

Steps to Solve

1. Resolve the 6N force into x and y components

The x-component is $6\cos(70^\circ)$ and the y-component is $6\sin(70^\circ)$.

2. Resolve the 5N force into x and y components

The x-component is $5\cos(160^\circ)$ and the y-component is $5\sin(160^\circ)$.

3. Resolve the 10N force into x and y components

The x-component is $10\cos(240^\circ)$ and the y-component is $10\sin(240^\circ)$.

4. Calculate the x-components

$6\cos(70^\circ) \approx 2.052$ $5\cos(160^\circ) \approx -4.698$ $10\cos(240^\circ) = -5$

5. Calculate the y-components

$6\sin(70^\circ) \approx 5.638$ $5\sin(160^\circ) \approx 1.710$ $10\sin(240^\circ) \approx -8.660$

6. Sum the x-components to find the resultant x-component $R_x = 2.052 - 4.698 - 5 = -7.646$

7. Sum the y-components to find the resultant y-component $R_y = 5.638 + 1.710 - 8.660 = -1.312$

8. Calculate the magnitude of the resultant force $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-7.646)^2 + (-1.312)^2} \approx \sqrt{58.46 + 1.72} \approx \sqrt{60.18} \approx 7.76$

9. Calculate the direction of the resultant force $\theta = \arctan\left(\frac{R_y}{R_x}\right) = \arctan\left(\frac{-1.312}{-7.646}\right) \approx \arctan(0.1716) \approx 9.74^\circ$

Since both $R_x$ and $R_y$ are negative, the angle is in the third quadrant. Therefore, we need to add 180 degrees to the arctan result. $\theta = 9.74^\circ + 180^\circ = 189.74^\circ$