Math Problem Statement

The figure here shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but that now moves across a frictionless floor. The force magnitudes are F1 = 2.80 N, F2 = 3.70 N, and F3 = 10.0 N, and the indicated angles are θ2 = 51.0° and θ3 = 33.0°. What is the net work done on the canister by the three forces during the first 3.50 m of displacement?

Solution

To solve this problem, we need to find the net work done on the cargo canister by the three forces during a displacement of 3.50 m. The forces are acting on the canister at given angles, and the formula for work is:

W=Fnetdcos(θ)W = F_{\text{net}} \cdot d \cdot \cos(\theta)

Where:

  • FnetF_{\text{net}} is the net force in the direction of the displacement,
  • dd is the displacement (3.50 m), and
  • θ\theta is the angle between the force and the displacement direction.

Steps:

  1. Decompose the forces: Since the forces are acting at angles, we need to find the components of each force in the horizontal (xx) and vertical (yy) directions.

    • Force F1=2.80NF_1 = 2.80 \, \text{N} is along the negative x-axis, so: F1x=2.80N,F1y=0NF_{1x} = -2.80 \, \text{N}, \quad F_{1y} = 0 \, \text{N}

    • Force F2=3.70NF_2 = 3.70 \, \text{N} is at an angle θ2=51.0\theta_2 = 51.0^\circ from the negative x-axis: F2x=3.70cos(51.0)2.33N,F2y=3.70sin(51.0)2.88NF_{2x} = -3.70 \cdot \cos(51.0^\circ) \approx -2.33 \, \text{N}, \quad F_{2y} = 3.70 \cdot \sin(51.0^\circ) \approx 2.88 \, \text{N}

    • Force F3=10.0NF_3 = 10.0 \, \text{N} is at an angle θ3=33.0\theta_3 = 33.0^\circ from the positive x-axis: F3x=10.0cos(33.0)8.38N,F3y=10.0sin(33.0)5.45NF_{3x} = 10.0 \cdot \cos(33.0^\circ) \approx 8.38 \, \text{N}, \quad F_{3y} = 10.0 \cdot \sin(33.0^\circ) \approx 5.45 \, \text{N}

  2. Calculate the net force in each direction: Fnet,x=F1x+F2x+F3x=2.802.33+8.383.25NF_{\text{net},x} = F_{1x} + F_{2x} + F_{3x} = -2.80 - 2.33 + 8.38 \approx 3.25 \, \text{N} Fnet,y=F1y+F2y+F3y=0+2.88+5.458.33NF_{\text{net},y} = F_{1y} + F_{2y} + F_{3y} = 0 + 2.88 + 5.45 \approx 8.33 \, \text{N}

  3. Find the magnitude of the net force: Fnet=(Fnet,x)2+(Fnet,y)2=(3.25)2+(8.33)28.92NF_{\text{net}} = \sqrt{(F_{\text{net},x})^2 + (F_{\text{net},y})^2} = \sqrt{(3.25)^2 + (8.33)^2} \approx 8.92 \, \text{N}

  4. Calculate the angle of the net force: θnet=tan1(Fnet,yFnet,x)=tan1(8.333.25)68.1\theta_{\text{net}} = \tan^{-1} \left( \frac{F_{\text{net},y}}{F_{\text{net},x}} \right) = \tan^{-1} \left( \frac{8.33}{3.25} \right) \approx 68.1^\circ

  5. Find the work done: Since the displacement is along the x-axis, we only need the horizontal component of the net force: W=Fnet,xd=3.253.5011.38JW = F_{\text{net},x} \cdot d = 3.25 \cdot 3.50 \approx 11.38 \, \text{J}

Thus, the net work done on the canister by the three forces is approximately 11.38 joules.

Would you like more details or have any questions?

Here are 5 related questions:

  1. What is the significance of decomposing forces into components?
  2. How do angles affect the calculation of work done by forces?
  3. What would happen if the displacement was in a different direction?
  4. How would you calculate the work done by individual forces?
  5. Can the net work be negative in any scenario?

Tip: Always remember to decompose vectors into components before applying them to different directions of motion!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Physics
Work
Vector Decomposition
Trigonometry

Formulas

Work formula: W = F_net * d * cos(θ)
Vector components: Fx = F * cos(θ), Fy = F * sin(θ)
Pythagorean theorem: F_net = √(Fx^2 + Fy^2)
Inverse tangent: θ_net = tan^(-1)(Fy / Fx)

Theorems

Pythagorean Theorem
Trigonometric Ratios

Suitable Grade Level

Grade 11-12