Question

Forces with magnitudes of v = 135 newtons and u = 280 newtons act on a hook (see figure). The angle between the two forces is 45°. Find the direction and magnitude of the resultant of these forces. (Hint: Write the vector representing each force in component form, then add the vectors. Round your answers to two decimal places.)

Answer 1

The magnitude of the resultant force is 387.41 N

The direction of the resultant force is 14.25⁰.

How to calculate the resultant of the vector?

The resultant of the vector v and vector u is calculated by applying the following method as shown below;

The given vectors include;

v = 135 N at angle 45⁰

u = 280 N at angle 0⁰

The x component of the vectors is calculated as follows;

vx = v cos(θ)

vx = 135 N x cos (45)

vx = 95.46 N

ux = u cos(θ)

ux = 280 N x cos(0)

ux = 280 N

x = 95.46 N + 280 N

x = 375.46 N

The y component of the vectors is calculated as follows;

vy = 135 x sin (45)

vy = 95.46 N

uy = 280 x sin (0)

uy = 0 N

y = 95.46 N

The magnitude of the resultant force is;

R = √ (375.46² + 95.46²)

R = 387.41 N

The direction of the resultant force is;

tan θ = y / x

tan θ = 95.46 / 375.46

tan θ = 0.254

θ = arc tan (0.254)

θ = 14.25⁰

Answer 2

To find the resultant force we will first find the component force of the forces given, to do this, we need to remember that any vector can be express in component form by:

`\begin{gathered} \vec{v}= \\ \text{ where } \\ v\text{ is the magnitude } \\ \theta\text{ is the angle of the vector with respect to the positive x-axis.} \end{gathered}`

For force u we notice that its magnitude is 280 N and its angle is zero, then we have:

`\begin{gathered} \vec{u}=<280\cos0,280\sin0> \\ \vec{u}=<280,0> \end{gathered}`

For force v we know that its magnitude is 135 N and its angle is 45°, then we have:

`\begin{gathered} \vec{v}=<135\cos45,135\sin45> \\ \vec{v}=<95.46,95.46> \end{gathered}`

Now that we have both vectors in component form we add them to get the resultant in component form:

`\begin{gathered} \vec{F}=\vec{u}+\vec{v} \\ \vec{F}=<280,0>+<95.46,95.46> \\ \vec{F}=<375.46,95.46> \end{gathered}`

Once we have the resultant force in component form we can find its magnitude and direction if we remember that they are given by:

`\begin{gathered} F=√(F_x+F_y) \\ \theta=\tan^(-1)((F_y)/(F_x)) \end{gathered}`

Plugging the values we found for the components we have:

`\begin{gathered} F=√(375.46^2+95.46^2) \\ F=387.41 \end{gathered}`

and

`\begin{gathered} \theta=\tan^(-1)((95.46)/(375.46)) \\ \theta=14.27 \end{gathered}`

Therefore, the magnitude of the resultant force is 387.41 N and the direction is 14.27°