Answer to Question #241011 in Physics for Bonbon

Question

Scalar and Vector Quantities


Solve the following problems:

1. Two forces A and B acting on a particle. Force A is 3 N directed north and B is 4 newton, 30 degrees of east. Find the resultant force using the laws of sines and cosine

2. Suppose we have a Resultant Force of 30 N in the direction 30 degrees North of East, find the x component and the y component

3 A jogger runs 4 m,40 degrees N of E, 2 m east, 5.20 m 30 degrees S of W, 6.50 m S, and then collapses. Find his resultant vector

4. To go to a nearby convenience store from her dormitory, student walk 12 meters north and 5 meter east. What is the students displacement from her dormitory?.

Answer 1

1. Apply the law of cosines to find the magnitude and the law of sines to find the angle toward East counting from North:

$$R=\sqrt{A^2+B^2-2AB\cos\rho}=\\ =\sqrt{3^2+4^2-2·3·4\cos120°}=6.1\text{ N}.\\\space\\ \frac{B}{\sin\beta}=\frac{R}{\sin\rho},\\\space\\ \beta=\arcsin\bigg(\frac BR\sin\rho\bigg)=34.6°\text{ E of N}.$$

2. The components are

$$R_x=R_E=30\cos30=26\text N,\\ R_y=R_N=30\sin30°=15\text N.$$

3. The resultant vector:

$$R_E=4\cos40°+2-5.2\cos30°-0=0.56\text m.\\ R_N=4\sin40°+2-5.2\sin30°-6.5=-4.53\text m.\\ R=\sqrt{R_N^2+R_E^2}=4.56\text{ m},\\\space\\ \theta=\arctan\frac{R_N}{R_E}=-83°\text{ S of E.}$$

4. The displacement is

$$D=\sqrt{15^2+5^2}=15.8\text{ m}.$$