Question

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant.A.What is the x component Fx of the resultant force?Express your answer in newtons.B. What is the y component Fy of the resultant force?Express your answer in newtons.C. What is the magnitude F of the resultant force?Express your answer in newtons.D. What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.Express your answer in degrees.

Answer 1

Answer:

$F{x}=-7.31\ N$

$F{y}=3.65\ N$

$|\vec F_n|=8.17\ N$

$\theta=26.53^o$

Explanation:

Net Force

When a set of forces is applied to an object, the net force is the vector sum of all the individual forces:

$\vec F_n=\vec F_1+\vec F_2+...+\vec F_k$

We have two forces described in the question: F1 has a magnitude of 9.2 N and is directed at $\theta_1=65^o$. The angle is positive since it's measured above the negative x-axis and clockwise, exactly as the problem requires the reference to be taken. F1 is expressed in vector form as:

$\vec F_1=\left< -9.2cos65^o,9.2sin65^o\right>$

$\vec F_1=\left< -3.89,8.34\right>\ N$

F2 has a magnitude of 5.8 N and is directed at $\theta_2=-53.9^o$. Expressing F2 in vector form:

$\vec F_2=\left< -5.8cos53.9^o,-5.8sin53.9^o\right>$

$\vec F_2=\left< -3.42,-4.69\right>\ N$

A. Let's compute the net or resultant force:

$\vec F_n=\left< -3.89,8.34\right>+\left< -3.42,-4.69\right>$

$\vec F_n=\left< -7.31,3.65\right>\ N$

The x-component of the resultant force is

$F{x}=-7.31\ N$

B. The y-component of the resultant force is

$F{y}=3.65\ N$

C. The magnitude of the resultant force is

$|\vec F_n|=\sqrt{(-7.31)^2+3.65^2}=8.17$

$|\vec F_n|=8.17\ N$

D. The angle of the resultant force is computed by

$\displaystyle tan\theta=\frac{3.65}{7.31}=0.5$

$\theta=26.53^o$

Since the x-component of the resultant force is negative and the y-component is positive, the angle lies in the second quadrant and is positive according to the required positive reference