At one moment in a football game, player A exerts a force to the east on player \(\mathrm{B}\). At the same time, a teammate of \(\mathrm{A}\) exerts the samesized force to the south on player \(\mathrm{B}\). In what direction is \(\mathrm{B}\) likely to go because of these forces?

Short Answer

Expert verified
Answer: Player B is likely to move in the direction that is 45 degrees south of east.

Step by step solution

01

Representing force vectors

Let the force exerted by player A to the east be represented by vector \(\vec{F_A}\) and the force exerted by A's teammate to the south be represented by vector \(\vec{F_B}\). Since both forces have the same magnitude, we can assume their magnitudes to be F. Then, \(\vec{F_A} = F\hat{i}\) (east direction) and \(\vec{F_B} = -F\hat{j}\) (south direction).
02

Calculating the resultant force

To find the resultant force acting on player B, we need to add the two force vectors \(\vec{F_A}\) and \(\vec{F_B}\). The resultant force \(\vec{R} = \vec{F_A} + \vec{F_B}\) can be calculated as follows: \(\vec{R} = F\hat{i} + (-F\hat{j}) = F\hat{i} - F\hat{j}\)
03

Finding the direction of the resultant force

To find the direction of the resultant force, we need to calculate the angle \(\theta\) between the resultant force vector and the x-axis (east). We can use the tangent function to calculate the angle: \(\tan{\theta} = \frac{|\vec{R_y}|}{|\vec{R_x}|}\) \(\tan{\theta} = \frac{F}{F}\) \(\tan{\theta} = 1\) The angle \(\theta\) can be calculated as: \(\theta = \tan^{-1}(1)\) \(\theta = 45^\circ\)
04

Finding the direction of player B's motion

Since the angle between the resultant force and the x-axis (east) is \(45^\circ\), player B is likely to move in the direction that is \(45^\circ\) south of east due to these forces.

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