Sure, let's analyze the scenario step by step.
### 1. Forces Acting Perpendicular to Each Other
When two forces act perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant force. Given:
- \( F_1 = 5 \, \text{N} \)
- \( F_2 = 12 \, \text{N} \)
The resultant force \( R \) can be calculated as:
\[ R = \sqrt{F_1^2 + F_2^2} \]
Plugging in the values:
\[ R = \sqrt{5^2 + 12^2} \]
\[ R = \sqrt{25 + 144} \]
\[ R = \sqrt{169} \]
\[ R = 13 \, \text{N} \]
So, the resultant force is \( 13 \, \text{N} \).
### 2. Forces Acting at an Angle of 40 Degrees to Each Other
When the forces act at an angle \(\theta\) to each other, we can use the law of cosines to find the magnitude of the resultant force. Given:
- \( F_1 = 5 \, \text{N} \)
- \( F_2 = 12 \, \text{N} \)
- \(\theta = 40^\circ\)
The resultant force \( R \) can be calculated as:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta} \]
Plugging in the values:
\[ R = \sqrt{5^2 + 12^2 + 2 \cdot 5 \cdot 12 \cdot \cos(40^\circ)} \]
\[ R = \sqrt{25 + 144 + 120 \cdot \cos(40^\circ)} \]
\[ R = \sqrt{169 + 120 \cdot \cos(40^\circ)} \]
Using \(\cos(40^\circ) \approx 0.766\):
\[ R = \sqrt{169 + 120 \cdot 0.766} \]
\[ R = \sqrt{169 + 91.92} \]
\[ R = \sqrt{260.92} \]
\[ R \approx 16.15 \, \text{N} \]
So, the resultant force when the forces act at an angle of 40 degrees to each other is approximately \( 16.15 \, \text{N} \).
An object of mass is acted upon by two forces of 5N and 12N. Calculate the resultant of the forces if:the forces act perpendicular to each other,, The two forces act at an angle of 40 degree to each other
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