Understanding forces acting at angles is crucial in fields like physics and engineering, where the application of force often involves multiple directions. The “4-7 Worksheet: Forces at Angles” presents scenarios where forces are applied at various angles, requiring students to calculate components of forces and solve problems related to equilibrium, vector addition, and resultant forces.
In this article, we will provide an answer key for the worksheet, offering step-by-step solutions to help students grasp the principles of forces at angles.
Understanding the Worksheet
The “4-7 Worksheet: Forces at Angles” typically covers concepts related to:
- Vector components: Breaking down forces into their horizontal and vertical components.
- Trigonometry: Using sine, cosine, and tangent functions to resolve forces.
- Resultant force: Combining components to find the overall effect of forces.
- Equilibrium: Applying Newton’s laws to determine if an object is in equilibrium when multiple forces are acting at different angles.
These types of problems help students to not only practice mathematical techniques like vector addition and trigonometry but also develop an intuitive understanding of how forces interact in real-world situations.
Answer Key: Solutions for “4-7 Worksheet: Forces at Angles”
Here is the solution breakdown for a typical “4-7 Worksheet: Forces at Angles.”
Problem 1: Calculating Force Components
Given:
- A force of 50 N is applied at an angle of 30° to the horizontal.
- Find the horizontal and vertical components of the force.
Solution: To solve for the horizontal and vertical components, we use trigonometric functions.
- Horizontal component:
Fx=F⋅cos(θ)=50⋅cos(30∘)≈50⋅0.866=43.3 NF_x = F \cdot \cos(\theta) = 50 \cdot \cos(30^\circ) \approx 50 \cdot 0.866 = 43.3 \, \text{N} - Vertical component:
Fy=F⋅sin(θ)=50⋅sin(30∘)≈50⋅0.5=25 NF_y = F \cdot \sin(\theta) = 50 \cdot \sin(30^\circ) \approx 50 \cdot 0.5 = 25 \, \text{N}
Thus, the horizontal component is 43.3 N, and the vertical component is 25 N.
Problem 2: Resultant Force from Two Forces at an Angle
Given:
- A 40 N force is applied at an angle of 45° to the horizontal.
- Another 60 N force is applied at an angle of 120° to the horizontal.
- Find the resultant force and its direction.
Solution: To solve this, we first resolve both forces into horizontal and vertical components.
For the 40 N force at 45°:
- Fx1=40⋅cos(45∘)=40⋅0.707≈28.3 NF_{x1} = 40 \cdot \cos(45^\circ) = 40 \cdot 0.707 \approx 28.3 \, \text{N}
- Fy1=40⋅sin(45∘)=40⋅0.707≈28.3 NF_{y1} = 40 \cdot \sin(45^\circ) = 40 \cdot 0.707 \approx 28.3 \, \text{N}
For the 60 N force at 120°:
- Fx2=60⋅cos(120∘)=60⋅(−0.5)=−30 NF_{x2} = 60 \cdot \cos(120^\circ) = 60 \cdot (-0.5) = -30 \, \text{N}
- Fy2=60⋅sin(120∘)=60⋅0.866≈51.96 NF_{y2} = 60 \cdot \sin(120^\circ) = 60 \cdot 0.866 \approx 51.96 \, \text{N}
Now, add the components to find the total force:
- Horizontal component of the resultant:
Fx=Fx1+Fx2=28.3+(−30)=−1.7 NF_{x} = F_{x1} + F_{x2} = 28.3 + (-30) = -1.7 \, \text{N} - Vertical component of the resultant:
Fy=Fy1+Fy2=28.3+51.96≈80.26 NF_{y} = F_{y1} + F_{y2} = 28.3 + 51.96 \approx 80.26 \, \text{N}
Finally, the resultant force magnitude is:
- R=Fx2+Fy2=(−1.7)2+(80.26)2≈2.89+6441.7=6444.59≈80.3 NR = \sqrt{F_x^2 + F_y^2} = \sqrt{(-1.7)^2 + (80.26)^2} \approx \sqrt{2.89 + 6441.7} = \sqrt{6444.59} \approx 80.3 \, \text{N}
To find the direction:
- θ=tan−1(FyFx)=tan−1(80.26−1.7)≈tan−1(−47.15)\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{80.26}{-1.7}\right) \approx \tan^{-1}(-47.15)
Since the angle is in the second quadrant, we calculate the angle as approximately 90° to 180°, giving:
- θ≈91.3∘\theta \approx 91.3^\circ from the positive x-axis.
Thus, the resultant force is approximately 80.3 N at an angle of 91.3° to the horizontal.
Problem 3: Object in Equilibrium
Given:
- A box is subjected to three forces: 30 N at 0°, 40 N at 90°, and 50 N at 180°.
- Determine if the box is in equilibrium.
Solution: For the box to be in equilibrium, the sum of the horizontal components and the sum of the vertical components must both be zero.
- Horizontal components:
Fx1=30⋅cos(0∘)=30⋅1=30 NF_{x1} = 30 \cdot \cos(0^\circ) = 30 \cdot 1 = 30 \, \text{N}
Fx2=40⋅cos(90∘)=0 NF_{x2} = 40 \cdot \cos(90^\circ) = 0 \, \text{N}
Fx3=50⋅cos(180∘)=50⋅(−1)=−50 NF_{x3} = 50 \cdot \cos(180^\circ) = 50 \cdot (-1) = -50 \, \text{N}Sum of horizontal components:
Fxtotal=30+0−50=−20 NF_{x\text{total}} = 30 + 0 – 50 = -20 \, \text{N} - Vertical components:
Fy1=30⋅sin(0∘)=0 NF_{y1} = 30 \cdot \sin(0^\circ) = 0 \, \text{N}
Fy2=40⋅sin(90∘)=40 NF_{y2} = 40 \cdot \sin(90^\circ) = 40 \, \text{N}
Fy3=50⋅sin(180∘)=0 NF_{y3} = 50 \cdot \sin(180^\circ) = 0 \, \text{N}Sum of vertical components:
Fytotal=0+40+0=40 NF_{y\text{total}} = 0 + 40 + 0 = 40 \, \text{N}
Since the horizontal and vertical components do not add up to zero (there is a net horizontal force of -20 N and a net vertical force of 40 N), the box is not in equilibrium.
Conclusion
The “4-7 Worksheet: Forces at Angles” challenges students to apply trigonometric principles to resolve forces into components and to calculate resultant forces. By carefully working through each problem, one can understand how forces interact at different angles and the importance of vector addition in analyzing complex systems. This answer key provides step-by-step explanations, ensuring that students gain a deeper understanding of the underlying physics principles.
Through continued practice, students will develop the ability to quickly resolve forces at angles, a fundamental skill in physics and engineering applications.
