A force of 5.0 n is applied to a point object at 30o with the horizontal. Another force of 10.0 n is applied at 120o . a) determine the net force (the magnitude and angle)
The Correct Answer and Explanation is:
To determine the net force acting on the point object, we will break both forces into their horizontal (x) and vertical (y) components and then sum them to find the resultant force. The forces are applied at angles with respect to the horizontal axis, so trigonometry is used to resolve them into components.
Given:
- Force 1 (F₁) = 5.0 N at 30° with the horizontal.
- Force 2 (F₂) = 10.0 N at 120° with the horizontal.
Step 1: Resolve each force into horizontal and vertical components.
For Force 1 (F₁ = 5.0 N at 30°):
- Horizontal component of F₁:
( F₁_x = F₁ \times \cos(30^\circ) = 5.0 \times \cos(30^\circ) = 5.0 \times 0.866 = 4.33 \, \text{N} ) - Vertical component of F₁:
( F₁_y = F₁ \times \sin(30^\circ) = 5.0 \times \sin(30^\circ) = 5.0 \times 0.5 = 2.5 \, \text{N} )
For Force 2 (F₂ = 10.0 N at 120°):
- Horizontal component of F₂:
( F₂_x = F₂ \times \cos(120^\circ) = 10.0 \times \cos(120^\circ) = 10.0 \times (-0.5) = -5.0 \, \text{N} ) - Vertical component of F₂:
( F₂_y = F₂ \times \sin(120^\circ) = 10.0 \times \sin(120^\circ) = 10.0 \times 0.866 = 8.66 \, \text{N} )
Step 2: Sum the components to find the resultant force.
Now we sum the horizontal and vertical components from both forces:
- Resultant horizontal component (Rₓ):
( Rₓ = F₁_x + F₂_x = 4.33 \, \text{N} + (-5.0 \, \text{N}) = -0.67 \, \text{N} ) - Resultant vertical component (Rᵧ):
( Rᵧ = F₁_y + F₂_y = 2.5 \, \text{N} + 8.66 \, \text{N} = 11.16 \, \text{N} )
Step 3: Find the magnitude and angle of the net force.
The magnitude of the net force (R) can be found using the Pythagorean theorem:
[
R = \sqrt{Rₓ^2 + Rᵧ^2} = \sqrt{(-0.67)^2 + (11.16)^2} = \sqrt{0.4489 + 124.5746} = \sqrt{125.0235} = 11.18 \, \text{N}
]
To find the angle (θ) of the net force, we use the tangent function:
[
\tan(\theta) = \frac{Rᵧ}{Rₓ} = \frac{11.16}{-0.67}
]
[
\theta = \tan^{-1}\left(\frac{11.16}{-0.67}\right) \approx \tan^{-1}(-16.7) \approx -86.7^\circ
]
Since the angle is negative and the horizontal component is to the left (in the negative x-direction), we adjust the angle to be measured from the positive x-axis in the second quadrant:
[
\theta = 180^\circ – 86.7^\circ = 93.3^\circ
]
Final Answer:
- The magnitude of the net force is 11.18 N.
- The angle of the net force with respect to the horizontal is 93.3°.
This solution involves resolving the forces into their components, summing them, and using trigonometry to find the magnitude and direction of the resultant force.