Question

A person is pushing a wheelbarrow along a ramp that makes an angle α = 33.0° with the horizontal. The wheelbarrow and load have a combined mass of 23.65 kg with the center of mass at the midpoint of the length L. What is the total force the person must apply so that the wheelbarrow remains horizontal on that ramp?

Answer 1

Answer:

$F = 138.2 N$

Explanation:

Since the wheelbarrow is maintained at horizontal position

so we can use Torque balance equation

$F_y L = mg\frac{L}{2}$

$F_y = \frac{mg}{2}$

$F_y = \frac{23.65 \times 9.8}{2}$

$F_y = 115.9 N$

Now force balance in Y direction we have

$F_y + F_n cos\theta = mg$

$\frac{mg}{2} + F_ncos33 = mg$

$F_n = \frac{mg}{2cos33}$

$F_n = \frac{23.65 \times 9.8}{2cos33}$

$F_n = 138.2 N$

Now force balance in x direction we have

$F_x = F_n sin\theta$

$F_x = 138.2 sin33$

$F_x = 75.25 N$

Now net force exerted by the person

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{75.25^2 + 115.9^2}$

$F = 138.2 N$