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A box of mass 75 kg lies on a rough surface inclined at an angle of 23° to the horizontal. The box is stationary on the incline as shown in the diagram below.

Physics

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Answer

287.27 N\text{287.27 N}

0.42\text{0.42}

Step-by-step solution

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3.1 A static frictional force is the force that opposes the tendency of motion of a stationary object relative to a surface it is in contact with. It acts parallel to the surface and prevents the object from sliding.

3.2
The free-body diagram for the box on the incline includes:

  • Gravitational force (FgF_g or mgmg) acting vertically downwards from the center of mass.
  • Normal force (NN) acting perpendicular to the surface, upwards from the surface.
  • Static frictional force (fsf_s) acting parallel to the surface, up the incline, opposing the tendency to slide down.
\begin{tikzpicture}[scale=0.8] % Incline \draw (0,0) -- (6,2.4); % Angle is atan(2.4/6) = 21.8 degrees, close enough to 23 \draw (0,0) -- (2,0); \draw (1.5,0) arc (0:23:1.5); \node at (1.8,0.3) {$23^\circ$}; % Box \draw[fill=lightgray] (2.5,1) rectangle (4.5,2); \node at (3.5,1.5) {75 kg}; % Forces % Gravity \draw[->, thick] (3.5,1.5) -- (3.5,0) node[below] {$F_g = mg$}; % Normal \draw[->, thick] (3.5,1.5) -- ({3.5 - 0.5*sin(23)}, {1.5 + 0.5*cos(23)}) node[above right] {$N$}; % Friction \draw[->, thick] (3.5,1.5) -- ({3.5 - 0.5*cos(23)}, {1.5 - 0.5*sin(23)}) node[below left] {$f_s$}; \end{tikzpicture}

3.3.1

  1. Identify the forces acting parallel to the incline.

    Since the box is stationary, the net force parallel to the incline is zero. The component of gravity acting down the incline is mgsin⁡θmg \sin\theta, and the static frictional force fsf_s acts up the incline.

    ∑Fparallel=0  ⟹  fs−mgsin⁡θ=0\sum F_{\text{parallel}} = 0 \implies f_s - mg \sin\theta = 0 fs=mgsin⁡θf_s = mg \sin\theta

  2. Substitute the given values (m=75 kgm = 75 \text{ kg}, g=9.8 m/s2g = 9.8 \text{ m/s}^2, θ=23∘\theta = 23^\circ).

    fs=(75 kg)(9.8 m/s2)sin⁡(23∘)f_s = (75 \text{ kg})(9.8 \text{ m/s}^2) \sin(23^\circ)

  3. Calculate the magnitude of the static frictional force.

    fs=735×0.3907f_s = 735 \times 0.3907 fs=287.2695 Nf_s = 287.2695 \text{ N} fs≈287.27 Nf_s \approx 287.27 \text{ N}

    The magnitude of the static frictional force is 287.27 N\boxed{\text{287.27 N}}.

    3.3.2

  4. The coefficient of static friction μs\mu_s is related to the maximum static frictional force fs,maxf_{s,max} and the normal force NN by fs,max=μsNf_{s,max} = \mu_s N.

    For the box to be stationary, the actual static friction fsf_s must be less than or equal to fs,maxf_{s,max}. If we assume the box is at the verge of slipping or we are calculating the minimum coefficient required to keep it stationary at this angle, then fs=fs,maxf_s = f_{s,max}.

  5. Calculate the normal force NN.

    The normal force balances the component of gravity perpendicular to the incline.

    ∑Fperpendicular=0  ⟹  N−mgcos⁡θ=0\sum F_{\text{perpendicular}} = 0 \implies N - mg \cos\theta = 0 N=mgcos⁡θN = mg \cos\theta N=(75 kg)(9.8 m/s2)cos⁡(23∘)N = (75 \text{ kg})(9.8 \text{ m/s}^2) \cos(23^\circ) N=735×0.9205N = 735 \times 0.9205 N=676.5675 NN = 676.5675 \text{ N}

  6. Calculate the coefficient of static friction using fs=μsNf_s = \mu_s N.

    μs=fsN\mu_s = \frac{f_s}{N} μs=287.2695 N676.5675 N\mu_s = \frac{287.2695 \text{ N}}{676.5675 \text{ N}} μs=0.4246\mu_s = 0.4246 μs≈0.42\mu_s \approx 0.42

    The coefficient of static friction is 0.42\boxed{\text{0.42}}.

    3.4
    The static frictional force required to keep the box stationary is fs=mgsin⁡θf_s = mg \sin\theta.
    When the angle is tilted to 18∘18^\circ, the new static frictional force fs′f_s' will be mgsin⁡(18∘)mg \sin(18^\circ).
    Since sin⁡(18∘)<sin⁡(23∘)\sin(18^\circ) < \sin(23^\circ), the new static frictional force will be less than the value calculated in question 3.3.1.

    The correct choice is LESS THAN.

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