Chapter 13, Problem 76A

To determine

Net force on a rock.

Answer to Problem 76A

  $5.0\times {10}^{1}N$

Explanation of Solution

Given:

Weight of rock $=54N$ ,

Net force $F_{Net}=46N$

Density of water, ${\rho }_{water}=1.00\times {10}^{3}kg/m^3$

Formula used: Net force on an object submerged in water can be obtained by:

  $F_{net,water}=F_g-F_{buoyant}$

  $F_{net,water}=F_g-{\rho }_{water}Vg$

  $F_{net,liquid}=F_g-2{\rho }_{water}Vg$

Or $V=\frac{F_g-F_{net,water}}{2{\rho }_{water}g}$

Substituting the value of V in the first equation:

  $F_{net,water}=F_g-{\rho }_{water}gV$

  $F_{net,water}=F_g-\left(\frac{1}{2}\right)(F_g-F_{net,liquid})$

  $F_{net,water}=\left(\frac{1}{2}\right)(F_g+F_{net,liquid})$

Calculation: Net force on a rock submerged in water is:

  $F_{net}=\left(\frac{1}{2}\right)\left(54N+46N\right)$

  $F_{net}=5.0\times {10}^{1}N$

Conclusion: Hence, there is $5.0\times {10}^{1}N$ net force on a submerged rock.