Question

A 21.3 kg box rests on a table. A 16.8 kg box is placed on top of the first box, as shown. Determine the total force in N that the table exerts on the first box.

Answer 1

A $21.3\mathrm{kg}$ box rests on a table. A $16.8\mathrm{kg}$ box is placed on top of the first box, as shown. Determine the total force in N that the table exerts on the first box.

Solution:

$\mathrm{m}_1 = 21.3\mathrm{kg} - \mathrm{mass}$ of the first box;

$\mathrm{m}_2 = 16.8 \mathrm{~kg} - \mathrm{mass}$ of the second box;

$\mathrm{N}_{1}$ - reaction force that acts on the first box;

$\mathrm{N}_2$ - reaction force that acts on the second box;

$\mathrm{P}_{2,1}$ - force that the second box exerts on the first box;

$\mathrm{F}$ - total force that the table exerts on the first box;

$\mathrm{P_{i,t}}$ - force that the first box exerts on the table;

Newton's third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction to that of the first body:

1 box and 2 box: $\vec{\mathrm{N}}_2 = -\vec{\mathrm{P}}_{2,1}$ ; $|\vec{\mathrm{N}}_2| = |\vec{\mathrm{P}}_{2,1}| = \mathrm{m}_2\mathrm{g}$

1 box and table: $\vec{\mathrm{P}}_{1,\mathrm{t}} = -\vec{\mathrm{F}}$ ; $|\vec{\mathrm{P}}_{1,\mathrm{t}}| = |\vec{\mathrm{F}}|$

Newton's second law for the first box:

$$\vec {\mathrm {F}} + \vec {\mathrm {P}} _ {2, 1} + \overline {{\mathrm {m} _ {1} \mathrm {g}}} = \vec {0}$$ $$y: F - P _ {2, 1} - m _ {1} g = 0$$ $$F - m _ {2} g - m _ {1} g = 0$$ $$F = g \left(m _ {1} + m _ {2}\right) = 9. 8 \frac {N}{k g} \cdot (2 1. 3 k g + 1 6. 8 k g) = 3 7 3 N$$

Answer: total force that the table exerts on the first box is 373N