Question

As you will see in a later chapter, forces are vector quantities, and the total force on an object is the vector sum of all forces acting on it.In the figure below, a force F1 of magnitude 6.40 units acts on a crate at the origin in a direction = 35.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the crate in the direction of the positive y-axis. Find graphically the magnitude and direction (in degrees counterclockwise from the +x-axis) of the resultant force F1 + F2. Two forces act on a crate. Force vector F1 acts up and right on the right side of the crate at an acute angle above the horizontal. Force vector F2 acts vertically upwards on the top side of the crate.magnitude unitsdirection ° counterclockwise from the +x-

Answer 1

The magnitude of the resultant force = 10.13 units

The direction of the resultant force = 58.84° counterclockwise from the +x axis

STEP - BY - STEP EXPLANATION

What to find?

• The magnitude of the resultant force.

,

• The direction of the resultant force.

Given:

`F_1=\hat{6.4\cos35}\hat{x}+\hat{6.4\sin 35y}\hat{}`
`=\hat{5.243x}+\hat{3.6709}y`

For F2

`\begin{gathered} F_2=\hat{0x}+\hat{5y} \\ \\ F_2=\hat{5y} \end{gathered}`

Net force

F = f₁ + f₂

=5.243 x + (3.6709 + 5)y

=5.243 x + 8.6709y

The graphical representation is

Hence,

`\begin{gathered} \text{Magnitude}=\sqrt[]{(5.243)^2+(8.6709)^2}=\sqrt[]{102.673} \\ \text{ =10.13 units} \end{gathered}`

Hence, the magnitude is 10.13 units.

To calculate the direction;

`\tan \theta=(8.6709)/(5.243)=1.6538`
`\theta=\tan ^(-1)(1.6538)`
`\theta=58.84^(^o)`