Question

A 125 kg box is sling down a 27 degree ramp at a constant velocity of 1.12 m/s. What is the friction force acting on it?

Answer 1

Answer on Question #48128 – Physics – Mechanics | Kinematics | Dynamics

1. A $125\,\mathrm{kg}$ box is sling down a 27 degree ramp at a constant velocity of $1.12\,\mathrm{m/s}$. What is the friction force acting on it?

where $\vec{F}$ is the friction force.

As the velocity of the box is constant, the acceleration equals to zero ($\vec{a} = 0$).

Then, write down this law in projectives:

$$\left\{ \begin{array}{l} m \cdot 0 = mg \sin \alpha - F \\ m \cdot 0 = -mg \cos \alpha + N \end{array} \right.$$

One can obtain the friction force from the first equation of the system: $\boxed{F = mg \sin \alpha}$.

Let check the dimension: $\left[F\right] = m \cdot \frac{m}{s^2} = N$.

Let evaluate the quantity: $F = 125 \cdot 9.81 \cdot \sin 27{}^{\circ} = 557\,(N)$.

Answer: $557\,N$.

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