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Three forces in the x-y plane act on a 4.40 kg mass: 14.90 N directed at 57 degrees, 8.40 N...

Question:

Three forces in the x-y plane act on a 4.40 kg mass: 14.90 N directed at 57 degrees, 8.40 N directed at 168 degrees, and 5.30 N directed at 195 degrees. All angles are measured from the positive x-axis, with positive angles in the Counter-Clockwise direction.

Calculate the magnitude of the acceleration.

Calculate the direction of the resultant force using the same sign convention as above (in degrees).

Resolution of Vectors:

Let we have a vector {eq}\vec A {/eq} that is acting at an angle {eq}\theta {/eq} with the x-axis. Then it can be resolved into two components according to the following expression:

  • {eq}\vec A = |A| ( cos \theta \hat i + sin \theta \hat j) {/eq}

This is where:

  • |A| is the magnitude of the vector.

Answer and Explanation:

The given forces are:

  • {eq}F_A =14.90 \ N {/eq} acting at an angle {eq}\theta_1 =57^o {/eq} from X-axis.
  • {eq}F_B = 8.40 \space N {/eq} acting at an angle {eq}\theta_2 = 168^\circ {/eq} from X-axis.
  • {eq}F_C = 5.30 \space N {/eq} acting at an angle {eq}\theta_3 = 195^\circ {/eq} from X-axis.


Now resolving the forces into the x and y components we have:

Resolving into components we have,

{eq}\begin{align*} \vec F_A &= F_{Ax} +F_{Ay}\\ & = |F_A| \cos \theta_1 \hat i +|F_A|\sin \theta_1 \hat j \\ & = 14.90 \times \cos 57^o \hat i +14.90 \times \sin 57^o \hat j\\ &=8.1 \hat i + 12.5 \hat j\\\\ \\ \vec F_B &= F_{Bx} +F_{By}\\ & = |F_B| \cos \theta_2 \hat i +|F_B|\sin \theta_2 \hat j \\ & = 8.40 \cos 168^\circ \hat i +8.40 \sin 168^\circ \hat j \\ &=-8.2 \hat i + 1.7 \hat j \\\\ \vec F_C &= F_{Cx} +F_{Cy}\\ & = |F_C| \cos \theta_3 \hat i +|F_C|\sin \theta_3 \hat j \\ & = 5.30 \cos 195^\circ \hat i +5.30 \sin 195^\circ \hat j \\ &=-5.1 \hat i - 1.4 \hat j \\\\ \end{align*} {/eq}


Thus we have resulting force of the forces given by

{eq}\begin{align*} \vec R &=\vec F_A +\vec F_B + \vec F_C \\ &=8.1 \hat i + 12.5 \hat j -8.2 \hat i + 1.7 \hat j -5.1 \hat i - 1.4 \hat j \\ &=- 5.2 \hat i + 12.8 \hat j \\\\ \text{Magnitude of the resultant is}\\ R &=\sqrt {(-5.2)^2 +12.8^2 } \ N \\ & = 13.8 \ N\\\\ \text{Angle made by the resultant from +X-axis is given by}\\ \theta & = 180^o - \tan ^{-1} \frac{12.8}{5.2}\\ &=112.1 ^o \end{align*} {/eq}


Learn more about this topic:

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Vector Components: The Magnitude of a Vector

from UExcel Physics: Study Guide & Test Prep

Chapter 2 / Lesson 8
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