(Solved): Please do part a and b 4. Vector Resolution: Given a force vector of \( F=(0.300 \mathrm{~kg}) \ti ...

Please do part a and b

4. Vector Resolution: Given a force vector of \( F=(0.300 \mathrm{~kg}) \times g=2.94 \mathrm{~N} \) at \( 60^{\circ} \), resolve the vector into its \( x \) - and \( y \)-components. That is, find the magnitudes of \( F_{x} \) and \( F_{y} \) such that \( F_{x} \) lies along the \( x \)-axis \( \left(90^{\circ}\right), \mathbf{F}_{y} \) lies along the \( y \)-axis \( \left(0^{\circ}\right) \), and \( \mathbf{F}_{\mathbf{x}}+\mathbf{F}_{\mathbf{y}}=\mathbf{F} \). Do this using the following procedures: a) Graphical. Draw a vector diagram to scale (again, fill up a half-sheet of paper...record the scale factor on the drawing). Draw the vector \( \mathrm{F} \), then its \( \mathrm{x} \) - and \( \mathrm{y} \)-components. Measure the magnitudes of \( F_{x} \) and \( F_{y} \) with a ruler. b) Analytical. Compute the magnitudes of \( F_{x} \) and \( F_{y} \) using the trigonometric relationships described in the Theory section. c) Experimental. Clamp pulleys at \( 90^{\circ} \) (the \( \mathrm{x} \)-axis), at \( 0^{\circ} \) (the \( \mathrm{y} \)-axis) and at \( 240^{\circ} \) (the opposite direction from \( 60^{\circ} \)...remember \( \left.\mathbf{E}=-\mathbf{R}\right) \). Place a total force of \( (0.300 \mathrm{~kg}) \times \mathrm{g}= \) \( 2.94 \mathrm{~N} \) on the \( 240^{\circ} \) pulley. This force is then the equilibrant of \( \mathrm{F}=2.94 \mathrm{~N} \) at \( 60^{\circ} \). Add masses to the \( 90^{\circ} \) and \( 0^{\circ} \) hangers until the system is in equilibrium. The \( 90^{\circ} \) and \( 0^{\circ} \) forces are then the \( F_{x} \) and \( F_{y} \) components of \( F \).