Source: High school physics (Chinese)
Problem Sets:
Problem
On the floor of an elevator that is accelerating upwards at $a_e = 2 \text{ m/s}^2$, there is an inclined wedge A. The wedge has a length of 5 m and a height of 3 m. An object B with a mass of $m_B = 3$ kg is placed on the inclined surface of A. Both A and B are at rest relative to the elevator.
- Find the magnitude of the normal force exerted by B on A.
- Find the magnitude of the friction force exerted by B on A.
[Q1] The magnitude of the normal force is given by $N = m_B(g + a_e)\cos\theta$.
$$N = 28.32 \text{ N}$$[Q2] The magnitude of the friction force is given by $f_s = m_B(g + a_e)\sin\theta$.
$$f_s = 21.24 \text{ N}$$We analyze the problem in the non-inertial reference frame of the elevator. Since the elevator accelerates upwards with $a_e$, we can model the system as being in an effective gravitational field where the acceleration due to gravity is $g_{eff} = g + a_e$. Object B is at rest in this frame, so it is in static equilibrium. We assume $g = 9.8 \text{ m/s}^2$.
First, determine the angle of inclination $\theta$ of the wedge. Given the length (hypotenuse) $L = 5$ m and height $h = 3$ m:
$$\sin\theta = \frac{h}{L} = \frac{3}{5}$$The adjacent side is $b = \sqrt{L^2 - h^2} = \sqrt{5^2 - 3^2} = 4$ m.
$$\cos\theta = \frac{b}{L} = \frac{4}{5}$$The effective gravitational force on object B is $F_{g,eff} = m_B(g+a_e)$ directed vertically downwards. We set up a coordinate system with the x-axis parallel to the incline (pointing up) and the y-axis perpendicular to the incline.
The forces acting on B are the normal force $N$, the static friction force $f_s$, and the effective gravitational force $F_{g,eff}$. Applying the equilibrium conditions ($\sum \vec{F} = 0$):
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Sum of forces in the y-direction (perpendicular to the incline):
$$ \sum F_y = N - F_{g,eff} \cos\theta = 0 $$ $$ N = F_{g,eff} \cos\theta = m_B(g + a_e)\cos\theta $$ -
Sum of forces in the x-direction (parallel to the incline): The component of the effective gravitational force pulling B down the incline is $F_{g,eff} \sin\theta$. To maintain equilibrium, the static friction force $f_s$ must act up the incline.
$$ \sum F_x = f_s - F_{g,eff} \sin\theta = 0 $$ $$ f_s = F_{g,eff} \sin\theta = m_B(g + a_e)\sin\theta $$
By Newton's third law, the forces exerted by B on A are equal in magnitude to the forces exerted by A on B.
[Q1] Magnitude of the normal force: Substitute the given values into the derived expression for $N$.
$$N = (3 \text{ kg})(9.8 \text{ m/s}^2 + 2 \text{ m/s}^2)\left(\frac{4}{5}\right)$$ $$N = (3 \text{ kg})(11.8 \text{ m/s}^2)(0.8) = 28.32 \text{ N}$$[Q2] Magnitude of the friction force: Substitute the given values into the derived expression for $f_s$.
$$f_s = (3 \text{ kg})(9.8 \text{ m/s}^2 + 2 \text{ m/s}^2)\left(\frac{3}{5}\right)$$ $$f_s = (3 \text{ kg})(11.8 \text{ m/s}^2)(0.6) = 21.24 \text{ N}$$