Question

A 20.0 kg box rests on a table. (a) What is the weight of the box and the normal force acting on it? (b) A 10.0 kg box is placed on top of the 20.0 kg box, as shown in Fig. 4-43. Determine the normal force that the table exerts on the 20.0 kg box and the normal force that the 20.0 kg box exerts on the 10.0 kg box.

Short answer

(a) The weight of the box is 196 N, and 196 N normal force acts on it.

(b) The normal force on the 10.0 kg box is 98 N, and the normal force on the 20.0 kg box is 294 N.

Step-by-step solution

Step 1. Given data and assumption

Given data:

The bigger mass is $M=20.0\mathrm{kg}$.

The smaller mass is $m=10.0\mathrm{kg}$.

Assumption:

For part (a), assume that only a 20.0 kg box is on the table.

Step 2. Calculation of the forces on the 20.0 kg individual box

Part (a)

The weight of the box is

$\begin{array}{c}W=mg\\ =\left(20.0\mathrm{kg}\right)\times \left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\\ =196\mathrm{N}\end{array}$

From Newton’s third law, the normal force on the 20.0 kg box is

$\begin{array}{c}N=mg\\ =196\mathrm{N}\end{array}$

Hence, the weight of the box is 196 N, and 196 N normal force acts on it.

Step 3. Calculation of the forces for the combined system

Part (b)

The normal force on the 10.0 kg box is

$\begin{array}{c}{N}_{10}=mg\\ =\left(10.0\mathrm{kg}\right)\times \left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\\ =98\mathrm{N}\end{array}$

The normal force on the 20.0 kg box is

$\begin{array}{c}{N}_{20}=mg+Mg\\ =\left(m+M\right)g\\ =\left(10.0\mathrm{kg}+20.0\mathrm{kg}\right)\times \left(9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\\ =294\mathrm{N}\end{array}$

Hence, the normal force on the 10.0 kg box is 98 N, and the normal force on the 20.0 kg box is 294 N.