Question

A 224 kg crate i pushed horizontally with a force of 710N. If the coefficient of friction is .25, calculate the acceleration of the crate.

Answer 1

A $224\mathrm{kg}$ crate i pushed horizontally with a force of 710N. If the coefficient of friction is .25, calculate the acceleration of the crate.

Solution.

Newton's second law in vector form:

$$m \vec {a} = \vec {F} + \vec {F _ {f}} + \vec {N} + m \vec {g};$$

Projection on OX:

$$m a _ {x} = F _ {x} - F _ {f x} + 0 + 0 = F - F _ {f};$$

Projection on OY:

$$m a _ {y} = 0 + 0 + N _ {y} - m g _ {y} = N - m g;$$

$a_{y} = 0$ - a crate don't move on OY.

$$0 = N - m g;$$ $$N = m g;$$

$a_{y} = 0$ ; then $a_{x} = a$ .

$$m a = F - F _ {f};$$ $$F _ {f} = \mu N = \mu m g;$$

$F_{f} = \mu mg$ - force of friction.

$$m a = F - \mu m g;$$ $$a = \frac {F - \mu m g}{m} = \frac {F}{m} - \mu g;$$ $$a = \frac {F}{m} - \mu g.$$ $$a = \frac {710}{224} - 0.25 \cdot 9.8 \approx 0.72 \left(\frac {m}{s ^ {2}}\right).$$

Answer: $a = 0.72\frac{m}{s^2}$.