Solve 3.1.1 and 3.1.2 and 3.1.3 on this picture

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Here are the solutions for questions 3.1.1, 3.1.2, and 3.1.3.

Given:
Mass m=10m = 10 kg
Angle of inclination θ=30∘\theta = 30^\circ
Applied force F=120F = 120 N (up the incline)
Coefficient of kinetic friction μk=0.2\mu_k = 0.2
Acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2

3.1.1. The magnitude of the normal force

  1. Identify the forces perpendicular to the inclined plane.

    The normal force NN acts perpendicular to the surface, balancing the component of the gravitational force perpendicular to the plane.

    N=mgcos⁡θN = mg \cos \theta

  2. Substitute the given values and calculate NN.

    N=(10 kg)(9.8 m/s2)cos⁡30∘N = (10 \text{ kg})(9.8 \text{ m/s}^2) \cos 30^\circ N=98 N×32N = 98 \text{ N} \times \frac{\sqrt{3}}{2} N=493 NN = 49\sqrt{3} \text{ N} N≈49×1.73205 NN \approx 49 \times 1.73205 \text{ N} N≈84.87 NN \approx 84.87 \text{ N}

    The magnitude of the normal force is 493 N\boxed{49\sqrt{3} \text{ N}} or approximately 84.87 N\boxed{84.87 \text{ N}}.

    3.1.2. The magnitude of the kinetic friction force

  3. Use the formula for kinetic friction force.

    The kinetic friction force fkf_k is given by the product of the coefficient of kinetic friction and the normal force.

    fk=μkNf_k = \mu_k N

  4. Substitute the values for μk\mu_k and NN (from 3.1.1) and calculate fkf_k.

    fk=0.2×(493 N)f_k = 0.2 \times (49\sqrt{3} \text{ N}) fk=9.83 Nf_k = 9.8\sqrt{3} \text{ N} fk≈0.2×84.87 Nf_k \approx 0.2 \times 84.87 \text{ N} fk≈16.97 Nf_k \approx 16.97 \text{ N}

    The magnitude of the kinetic friction force is 9.83 N\boxed{9.8\sqrt{3} \text{ N}} or approximately 16.97 N\boxed{16.97 \text{ N}}.

    3.1.3. The magnitude of the net force

  5. Identify all forces acting parallel to the inclined plane.

    • Applied force F=120F = 120 N (up the incline)
    • Component of gravitational force parallel to the incline mgsin⁡θmg \sin \theta (down the incline)
    • Kinetic friction force fkf_k (down the incline, as the applied force FF is greater than mgsin⁡θmg \sin \theta, indicating upward motion or tendency to move upward).

  6. Calculate the component of gravitational force parallel to the incline.

    mgsin⁡θ=(10 kg)(9.8 m/s2)sin⁡30∘mg \sin \theta = (10 \text{ kg})(9.8 \text{ m/s}^2) \sin 30^\circ mgsin⁡θ=98 N×12mg \sin \theta = 98 \text{ N} \times \frac{1}{2} mgsin⁡θ=49 Nmg \sin \theta = 49 \text{ N}

  7. Calculate the net force by summing the forces parallel to the incline.

    Taking the upward direction along the incline as positive:

    Fnet=F−mgsin⁡θ−fkF_{net} = F - mg \sin \theta - f_k Fnet=120 N−49 N−9.83 NF_{net} = 120 \text{ N} - 49 \text{ N} - 9.8\sqrt{3} \text{ N} Fnet=71 N−9.83 NF_{net} = 71 \text{ N} - 9.8\sqrt{3} \text{ N} Fnet≈71 N−16.97 NF_{net} \approx 71 \text{ N} - 16.97 \text{ N} Fnet≈54.03 NF_{net} \approx 54.03 \text{ N}

    The magnitude of the net force is (71−9.83) N\boxed{(71 - 9.8\sqrt{3}) \text{ N}} or approximately 54.03 N\boxed{54.03 \text{ N}}.

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