\[ F_a = 4000\ N \]
– The angle between Fa and line L is $\theta_a = 45^{\circ}$.
– The angle between Fb and line L is $\theta_b = 35^{\circ}$.
The question aims to find the 2nd force exerted on the housing unit by a snowcat in Antarctica, and the sum of both forces’ magnitude exerted on the housing unit.
The question depends on the concept of Force, and two forces exerted on an object at an angle, and the resultant force. The force is a vector quantity; thus, it has a direction along with the magnitude. The resultant force is the vector sum of two forces acting on an object at different angles. The resultant force is given as:
\[ \overrightarrow{R} = \overrightarrow{F_1} + \overrightarrow{F_2} \]
Expert Answer
The sum of forces exerted by the snowcats on the housing unit is parallel to line L. This means that the forces must be balanced in the horizontal component. The balanced equation of the horizontal components of these forces is given as:
\[ F_a \cos \theta_a = F_b \cos \theta_b \]
Substituting the values, we get:
\[ 4000 \cos (45 ^{\circ}) = F_b \cos (35^ {\circ}) \]
Rearranging for $F_b$, we get:
\[ F_b = \dfrac{ 4000 \cos( 45^{\circ}) }{ \cos ( 35^{\circ} } \]
\[ F_b = \dfrac{ 4000 \times 0.707 }{ 0.819 } \]
\[ F_b = \dfrac{ 2828 }{ 0.819 } \]
\[ F_b = 3453\ N \]
The sum of both forces $F_a$ and $F_b$ is given as:
\[ \overrightarrow{F}^2 = \overrightarrow{F_a}^2 + \overrightarrow{F_b}^2 \]
The magnitude of $F_a$ is given as:
\[ F_a = 4000 \sin (45) \]
\[ F_a = 4000 \times 0.707 \]
\[ F_a = 2828\ N \]
The magnitude of $F_b$ is given as:
\[ F_b = 3453 \sin (35) \]
\[ F_b = 3453 \times 0.5736 \]
\[ F_b = 1981\ N \]
The sum of the magnitude of both forces is given as:
\[ F = \sqrt{ F_a^2 + F_b^2 } \]
Substituting the values, we get:
\[ F = \sqrt{ 2828^2 + 1981^2 } \]
\[ F = 3453\ N \]
Numerical Result
The magnitude of $F_b$ is calculated to be:
\[ F_b = 3453\ N \]
The magnitude of the sum of both forces is calculated to be:
\[ F = 3453\ N \]
Example
Two forces, 10N and 15N, are exerted on an object at an angle of 45. Find the resultant force on the object.
\[ F_a = 10\ N \]
\[ F_b = 15\ N \]
\[ \theta = 45^ {\circ} \]
The resultant force between these two forces is given as:
\[ F = \sqrt{ |F_a|^2 + |F_b|^2 } \]
The magnitude of $F_a$ is given as:
\[ F_a = 10 \sin (45) \]
\[ F_a = 10 \times 0.707 \]
\[ F_a = 7.07\ N \]
The magnitude of $F_b$ is given as:
\[ F_b = 15 \sin (45) \]
\[ F_b = 15 \times 0.707 \]
\[ F_b = 10.6\ N \]
The resultant force is given as:
\[ F = \sqrt{ 7.07^2 + 10.6^2 } \]
\[ F = \sqrt{ 49.98 + 112.36 } \]
\[ F = \sqrt{ 162.34 } \]
\[ F = 12.74\ N \]