91Ó°ÊÓ

When studying forces acting on an object, it's essential to understand the concept of net force. Net force is the overall force resulting from the combination of all individual forces acting on an object. According to Newton's Second Law, it is described by the equation \( \vec{F}_{net} = m \cdot \vec{a} \), where \( m \) is the mass of the object, and \( \vec{a} \) is its acceleration.

To find the net force, you need to account for all the forces involved. Begin by identifying all the forces acting on the object. Once the individual forces are identified, their vector components are added together to yield the net force.

In the given exercise, a 2.00 kg object is subjected to an acceleration of \((-8.00 \, \text{m/s}^2) \hat{i} + (6.00 \, \text{m/s}^2) \hat{j}\). The net force on the object can be found using the equation:

\[ \vec{F}_{net} = 2.00 \, \text{kg} \times \left( (-8.00 \, \text{m/s}^2) \hat{i} + (6.00 \, \text{m/s}^2) \hat{j} \right) = (-16.0 \, \text{N}) \hat{i} + (12.0 \, \text{N}) \hat{j}.\]

The forces in this exercise are presented as vectors, expressed with both magnitude and direction. In vector notation, individual components in different directions can be represented separately using unit vectors—\(\hat{i}\) for the x-direction and \(\hat{j}\) for the y-direction.

Understanding vector components allows us to work with each direction individually. For instance, if you have a force \( \vec{F}_{1} = (30.0 \, \text{N}) \hat{i} + (16.0 \, \text{N}) \hat{j} \), the vector components are:

• \(30.0 \, \text{N}\) along the x-axis (\(\hat{i}\)).
• \(16.0 \, \text{N}\) along the y-axis (\(\hat{j}\)).

Every force can be dissected similarly, focusing on its impact in each dimension separately, thereby simplifying the calculations as you sum or subtract these components individually to determine the net force.

Force calculation involves understanding how individual forces work together or against each other, impacting the motion of an object. Using the information from net force and vector components, you can determine unknown forces by rearranging known equations.

In the exercise, you want to find the third unknown force, \(\vec{F}_{3}\). The relationship given by Newton's law is:\[ \vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3}. \]
To solve for \(\vec{F}_{3}\), you rearrange this equation to:

\( \vec{F}_{3} = \vec{F}_{net} - \vec{F}_{1} - \vec{F}_{2}.\)

By substituting the values, \(\vec{F}_{net} = (-16.0 \, \text{N}) \hat{i} + (12.0 \, \text{N}) \hat{j}, \vec{F}_{1} = (30.0 \, \text{N}) \hat{i} + (16.0 \, \text{N}) \hat{j}, \) and \(\vec{F}_{2} = (-12.0 \, \text{N}) \hat{i} + (8.00 \, \text{N}) \hat{j},\) you can compute:

• For the x-component: \[-16.0 \, \text{N} - (30.0 \, \text{N}) + (-12.0 \, \text{N}) = -34.0 \, \text{N}\]
• For the y-component: \[12.0 \, \text{N} - (16.0 \, \text{N}) - (8.0 \, \text{N}) = -12.0 \, \text{N}\]

Thus, the third force \(\vec{F}_{3} = (-34.0 \, \text{N}) \hat{i} + (-12.0 \, \text{N}) \hat{j}.\) By decomposing and calculating forces this way, you gain a clear understanding of how different forces act on an object to produce motion.