Question

question completion status:1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 2021 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 4041 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60two displacement vectors, \\(\vec{a}\\) and \\(\vec{b}\\), are shown in the figure. the magnitudes of the displacements are \\(a = 10.0\\,\text{m}\\) and \\(b = 5.00\\,\text{m}\\).what is the \\(y\\)-component of resultant \\(\vec{r}\\), such that \\(\vec{r} = \vec{a} + \vec{b}\\)?hint: review section 1.8 in textbook (component method of vector addition), example 9 page 16 in textbook, and example 1-2 from our lecture video.\\(\

$$\begin{array}{c}\ \\hspace{3cm}\\vec{a}\\\ \\hspace{2cm}-x\\hspace{2cm}+x\\\ \\hspace{3cm}\\vec{b}\\\ \\hspace{4cm}-y\ \\end{array}$$

\\) (with angles: \\(\vec{a}\\) at \\(42.0^\circ\\) from +y towards -x, \\(\vec{b}\\) at \\(31.0^\circ\\) from -x towards -y)options: \\(-7.14\\,\text{m}\\), \\(-2.47\\,\text{m}\\), \\(6.26\\,\text{m}\\), \\(4.85\\,\text{m}\\)

Explanation:

Step1: Find y - component of \(\vec{A}\)

The vector \(\vec{A}\) makes an angle of \(42.0^{\circ}\) with the \(+y\) axis (towards the \(-x\) direction). The y - component of a vector \(\vec{V}\) with magnitude \(V\) and angle \(\theta\) (with respect to the y - axis) is given by \(V_y=V\cos\theta\) (since it is in the positive y - direction). For \(\vec{A}\), \(A = 10.0\space m\) and \(\theta=42.0^{\circ}\), so \(A_y=A\cos(42.0^{\circ})\).
\(A_y = 10.0\times\cos(42.0^{\circ})\approx10.0\times0.7431 = 7.431\space m\)

Step2: Find y - component of \(\vec{B}\)

The vector \(\vec{B}\) makes an angle of \(31.0^{\circ}\) with the \(-x\) axis (towards the \(-y\) direction). The angle of \(\vec{B}\) with the \(-y\) axis? Wait, let's see the direction. The vector \(\vec{B}\) is below the \(-x\) axis? No, the angle is \(31.0^{\circ}\) from the \(-x\) axis towards the \(-y\) axis? Wait, the standard way: to find the y - component, we can see that the angle between \(\vec{B}\) and the \(-x\) axis is \(31.0^{\circ}\), so the angle with the \(-y\) axis? Wait, no. Let's consider the coordinate system. The y - component of \(\vec{B}\): the vector \(\vec{B}\) is in the third quadrant? Wait, no, the \(-x\) and \(-y\) directions? Wait, the angle is \(31.0^{\circ}\) from the \(-x\) axis towards the \(-y\) axis. So the angle of \(\vec{B}\) with the \(x\) - axis (negative x - axis) is \(31.0^{\circ}\), so the angle with the \(y\) - axis (negative y - axis) would be \(90 - 31=59^{\circ}\)? No, better to use the formula for y - component. The y - component of a vector with magnitude \(B\) and angle \(\alpha\) (where \(\alpha\) is the angle between the vector and the \(-x\) axis, measured towards the \(-y\) direction) is \(B_y=-B\sin(31.0^{\circ})\) (negative because it is in the negative y - direction). For \(\vec{B}\), \(B = 5.00\space m\) and the angle for y - component: since the vector is below the \(-x\) axis, the y - component is negative. The angle between \(\vec{B}\) and the \(-x\) axis is \(31.0^{\circ}\), so the angle with the \(x\) - axis (if we take the standard position, from \(+x\) axis counter - clockwise) would be \(180^{\circ}- 31.0^{\circ}=149^{\circ}\)? No, maybe easier: the y - component of \(\vec{B}\) is \(B_y=-B\sin(31.0^{\circ})\) (because the vector is going downwards in y - direction). So \(B_y=5.00\times(-\sin(31.0^{\circ}))\)
\(\sin(31.0^{\circ})\approx0.5150\), so \(B_y=5.00\times(- 0.5150)=- 2.575\space m\)

Step3: Find y - component of \(\vec{R}\)

Since \(\vec{R}=\vec{A}+\vec{B}\), the y - component of \(\vec{R}\) is \(R_y = A_y + B_y\)
\(R_y=7.431-2.575 = 4.856\approx4.85\space m\)

Answer:

\(4.85\space m\)