Question

Sand drops at the rate of 2000 kg/min. from the bottom of a hopper onto a belt coveyor moving horizontally at 250 m/min. determine the force needed to drive the conveyor, neglecting friction?

a)500 N

b)800 N

c)139 N

d)152 N

Answer 1

Answer on Question #59419, Physics, Mechanics | Kinematics | Dynamics

Sand drops at the rate of $2000\,\mathrm{kg/min}$ from the bottom of a hopper onto a belt conveyor moving horizontally at $250\,\mathrm{m/min}$. Determine the force needed to drive the conveyor, neglecting friction.

Solution:

The initial horizontal velocity of the sand is zero.

The final horizontal velocity is $250\,\mathrm{m/min} = 250/60\,\mathrm{m/s}$.

The impulse of force is equal to the change in momentum of an object provided the mass is constant:

$$Impulse = F\Delta t = m\Delta v$$

Thus, the force is

$$F = \frac{m\Delta v}{\Delta t}$$ $$\frac{m}{\Delta t} = 2000\,\frac{\mathrm{kg}}{\mathrm{min}} = \frac{2000}{60}\,\frac{\mathrm{kg}}{\mathrm{s}}$$

The momentum change per second is

$$\frac{m\Delta v}{\Delta t} = \frac{2000 \cdot (250 - 0)}{60 \cdot 60} = 138.9\,\frac{\mathrm{kg}\,\mathrm{m}}{\mathrm{s}^2}.$$

Thus,

$$F = 138.9\,\mathrm{N}.$$

Answer. $F = 138.9\,\mathrm{N}$

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