Question

question 16
2 points save answer
(problem reference m.1) four displacement vectors are shown in the figure. the magnitudes of the displacements are: a = 2.25 cm, b = 6.35 cm, c
= 5.47 cm, and d = 4.19 cm.
what is the magnitude of the resultant?
6.98 cm
3.48 cm
16.9 cm
12.1 cm
18.2 cm

Explanation:

Step1: Calculate \(x\) - components of vectors

For vector \(\vec{A}\): \(A_x = A\cos(52.0^{\circ})=2.25\cos(52.0^{\circ})\approx2.25\times0.616 = 1.386\)
For vector \(\vec{B}\): \(B_x = B\cos(20.0^{\circ})=6.35\cos(20.0^{\circ})\approx6.35\times0.940 = 5.969\)
For vector \(\vec{C}\): \(C_x = C\sin(27.0^{\circ})=5.47\sin(27.0^{\circ})\approx5.47\times0.454 = 2.483\)
For vector \(\vec{D}\): \(D_x = - D\cos(31.0^{\circ})=- 4.19\cos(31.0^{\circ})\approx-4.19\times0.857=-3.591\)
Sum of \(x\) - components: \(R_x=A_x + B_x+C_x + D_x=1.386 + 5.969+2.483-3.591 = 6.247\)

Step2: Calculate \(y\) - components of vectors

For vector \(\vec{A}\): \(A_y = A\sin(52.0^{\circ})=2.25\sin(52.0^{\circ})\approx2.25\times0.788 = 1.773\)
For vector \(\vec{B}\): \(B_y = - B\sin(20.0^{\circ})=-6.35\sin(20.0^{\circ})\approx-6.35\times0.342=-2.172\)
For vector \(\vec{C}\): \(C_y = - C\cos(27.0^{\circ})=-5.47\cos(27.0^{\circ})\approx-5.47\times0.891=-4.874\)
For vector \(\vec{D}\): \(D_y = D\sin(31.0^{\circ})=4.19\sin(31.0^{\circ})\approx4.19\times0.515 = 2.168\)
Sum of \(y\) - components: \(R_y=A_y + B_y+C_y + D_y=1.773-2.172 - 4.874+2.168=-3.105\)

Step3: Calculate magnitude of resultant

Using \(R=\sqrt{R_x^{2}+R_y^{2}}\), \(R=\sqrt{(6.247)^{2}+(-3.105)^{2}}=\sqrt{39.035 + 9.641}=\sqrt{48.676}\approx6.98\)

Answer:

\(6.98\ cm\)