Chapter 1, Problem 24RQ

To determine

The weight of a 1 kg substance in N.

The weight of a 1 kg substance in kN.

The weight of a 1 kg substance in $\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$.

The weight of a 1 kg substance in kgf.

The weight of a 1 kg substance in $\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$.

The weight of a 1 kg substance in lbf.

Explanation of Solution

Given:

The mass of the substance $\left(m\right)$ is 1 kg.

The acceleration of gravity $\left(g\right)$  is

Calculation:

Write the expression of Newton’s second law for the weight of a substance.

  $W=mg$        (I)

For N,

Substitute $1\text{kg}$ for $\text{m}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(1\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.81\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =9.81\text{N}\end{array}$.

Thus, The weight of a 1 kg substance in N is $\underset{\_}{9.81\text{N}}$.

For kN,

Substitute $1\text{kg}$ for $\text{m}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(1\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.81\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \left(\frac{1\text{kN}}{1000\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =0.00981\text{kN}\end{array}$.

Thus, The weight of a 1 kg substance in kN is $\underset{\_}{0.00981\text{kN}}$.

For $\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$,

Substitute $1\text{kg}$ for $\text{m}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(1\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.81\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\end{array}$.

Thus, The weight of a 1 kg substance in $\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$ is $\underset{\_}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}$.

For kgf,

Substitute $1\text{kg}$ for $\text{m}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(1\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.81\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kgf}}{9.81\text{N}}\right)\\ =1\text{kgf}\end{array}$.

Thus, The weight of a 1 kg substance in kgf is $\underset{\_}{1\text{kgf}}$.

For $\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$,

Substitute $1\text{kg}$ for $\text{m}$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(1\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(1\text{kg}\right)\times \left(\frac{2.205\text{lbm}}{1\text{kg}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\\ =71\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}\end{array}$.

Thus, The weight of a 1 kg substance in $\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$ is $\underset{\_}{71\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}$.

For lbf,

Substitute $1\text{kg}$ for $\text{m}$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\begin{array}{c}W=\left(1\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(1\text{kg}\right)\times \left(\frac{2.205\text{lbm}}{1\text{kg}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\\ =2.21\text{lbf}\end{array}$.

Thus, The weight of a 1 kg substance in lbf is $\underset{\_}{2.21\text{lbf}}$.