Math Problem Statement

A horizontal force, F1 = 55 N, and a force, F2 = 12.8 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 3.1 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.

Randomized Variables F1 = 55 N F2 = 12.8 N m = 3.1 kg

  Part (a)

Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°. FN

Solution

To solve for the magnitude of the normal force FNF_N acting on the block, we will analyze the forces acting on it. The normal force FNF_N is the force exerted by the surface to balance the vertical forces acting on the block.

Forces acting on the block:

  1. The weight of the block, W=mgW = mg, acts downward, where: W=mg=3.1kg9.8m/s2W = m \cdot g = 3.1 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 This is the gravitational force.

  2. The force F2F_2 has both vertical and horizontal components due to the angle θ=30\theta = 30^\circ:

    • The vertical component of F2F_2 is F2sinθF_2 \sin \theta, which acts upward and reduces the normal force.
  3. The normal force FNF_N, which acts upward, opposes the net vertical force to prevent the block from moving in the vertical direction.

Calculation of FNF_N:

The normal force balances the net vertical forces. This gives us the following equation: FN=WF2sinθF_N = W - F_2 \sin \theta Where:

  • W=mgW = mg is the weight of the block,
  • F2sinθF_2 \sin \theta is the upward vertical component of force F2F_2.

Now, let’s calculate this step-by-step:

  1. Compute the weight of the block: W=3.1kg9.8m/s2=30.38NW = 3.1 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 30.38 \, \text{N}

  2. Calculate the vertical component of F2F_2: F2sin30=12.8Nsin30=12.8N0.5=6.4NF_2 \sin 30^\circ = 12.8 \, \text{N} \cdot \sin 30^\circ = 12.8 \, \text{N} \cdot 0.5 = 6.4 \, \text{N}

  3. Substitute into the normal force equation: FN=30.38N6.4N=23.98NF_N = 30.38 \, \text{N} - 6.4 \, \text{N} = 23.98 \, \text{N}

Thus, the magnitude of the normal force FNF_N is approximately: FN23.98NF_N \approx 23.98 \, \text{N}

Would you like more details on any step, or have any questions?

Here are 5 related questions for further exploration:

  1. What is the frictional force acting on the block?
  2. How would the normal force change if θ\theta were increased?
  3. What is the net horizontal force acting on the block?
  4. How does the coefficient of kinetic friction affect the motion of the block?
  5. How would the normal force be affected if the mass of the block increased?

Tip: The vertical component of a force at an angle can significantly reduce the normal force, thus affecting the friction.

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Math Problem Analysis

Mathematical Concepts

Physics
Forces
Trigonometry
Friction

Formulas

Weight: W = mg
Vertical component of force: F2 * sin(θ)
Normal force: FN = W - F2 * sin(θ)

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12