A cube 0.50 m on a side is made of bronze having a specific weight of 86.9 kN/m^3. Determine the magnitude and direction of the force required to hold the cube in equilibrium when completely submerged (a) in water and (b) in mercury. The specific gravity of mercury is 13.54.

Question

A cube 0.50 m on a side is made of bronze having a specific weight of 86.9[latex]{ kN}/{m^3}.[/latex] Determine the magnitude and direction of the force required to hold the cube in equilibrium when completely submerged (a) in water and (b) in mercury. The specific gravity of mercury is 13.54.

Accepted Answer

Consider part (a) first. Imagine the cube of bronze submerged in water. Now do Step 1 of the procedure.

On the assumption that the bronze cube will not stay in equilibrium by itself, some external force is required. The objective is to find the magnitude of this force and the direction in which it would act— that is, up or down.

Now do Step 2 of the procedure before looking at the next panel.

The free body is simply the cube itself. There are three forces acting on the cube in the vertical direction, as shown in Fig. 5.3:

[latex]\blacksquare [/latex] The weight of the cube w, acting downward through its center of gravity

[latex]\blacksquare [/latex] The buoyant force [latex]F_b[/latex], acting upward through the centroid of the displaced volume

[latex]\blacksquare [/latex] The externally applied supporting force [latex]F_e[/latex]

Part (a) of Fig. 5.3 shows the cube as a three-dimensional object with the three forces acting along a vertical line through the centroid of the volume. This is the preferred visualization of the free-body diagram. However, for most problems it is suitable to use a simplified two-dimensional sketch as shown in part (b).

How do we know to draw the force [latex]F_e[/latex] in the upward direction?

We really do not know for certain. However, experience should indicate that without an external force the solid bronze cube would tend to sink in water. Therefore, an upward force seems to be required to hold the cube in equilibrium. If our choice is wrong, the final result will indicate that to us.

Now, assuming that the forces are as shown in Fig. 5.3, go on to Step 3.

The equation should look as follows (assume that positive forces act upward):

[latex]\sum{F_v} =0[/latex]

[latex]F_b+F_e-W=0[/latex] (5-2)

As a part of Step 4, solve this equation algebraically for the desired term.

You should now have

[latex]F_e=W-F_b[/latex] (5-2)

because the objective is to find the external force.
How do we calculate the weight of the cube w?

Item b under Step 4 of the procedure indicates that [latex]W = \gamma_B V,[/latex] where [latex] \gamma_B [/latex]is the specific weight of the bronze cube and V is its total volume. For the cube, because each side is 0.50 m, we have

[latex]V = \left(0.50m\right)^3 = 0.125m^3 [/latex]

and

[latex]W = \gamma_B V= \left(86.9{kN}/{m^3}\right) \left(0.125m^3\right) =10.86kN[/latex]

There is another unknown on the right side of Eq. (5–3). How do we calculate [latex]F_b[/latex]?

Check Step 4a of the procedure if you have forgotten. Write

[latex]F_b = \gamma_f V_d[/latex]

In this case [latex] \gamma_f [/latex]is the specific weight of the water [latex]\left(9.81{kN}/{m^3}\right) [/latex], and the displaced volume [latex] V_d [/latex] is equal to the total volume of the cube, which we already know to be [latex] 0.125m^3[/latex]. Then, we have

[latex]F_b = \gamma_f V_d = \left(9.81{kN}/{m^3}\right) \left(0.125m^3\right) =1.23kN[/latex]

Now we can complete our solution for [latex] F_e [/latex].

The solution is

[latex]F_e = W - F_b = 10.86kN-1.23kN = 9.63kN[/latex]

Result Part (a) Notice that the result is positive. This means that our assumed direction for [latex]F_e[/latex] was correct. Then the solution to the problem is that an upward force of 9.63 kN is required to hold the block of bronze in equilibrium under water.

What about part (b) of the problem, where the cube is submerged in mercury? Our objective is the same as before—to determine the magnitude and direction of the force required to hold the cube in equilibrium.

Now do Step 2 of the procedure.

Either of the two free-body diagrams is correct as shown in Fig. 5.4, depending on the assumed direction for the external force [latex]F_e[/latex]. The solution for the two diagrams will be carried out simultaneously so you can check your work regardless of which diagram looks like yours, and to demonstrate that either approach will yield the correct answer.

Now do Step 3 of the procedure.

The following are the correct equations of equilibrium. Notice the differences and relate them to the figures:

[latex]F_b+F_e - W = 0 \mid F_b-F_e-W=0[/latex]

Now, solve for [latex]F_e[/latex].

You should now have

[latex]F_e=W - F_b \mid F_e=F_b-W[/latex]

Because the magnitudes of w and [latex]F_b[/latex] are the same for each equation, they can now be calculated.

As in part (a) of the problem, the weight of the cube is

[latex]W= \gamma_B V = \left(86.9{kN}/{m^3}\right) \left(0.125m^3\right) =10.86kN[/latex]

For the buoyant force [latex]F_b[/latex], you should have

[latex]F_b = \gamma_m V=\left(sg\right)_m \left(9.81{kN}/{m^3}\right) \left(V\right) [/latex]

where the subscript m refers to mercury. We then have

[latex]F_b =\left(13.45\right) \left(9.81{kN}/{m^3}\right) \left(0.125m^3\right) =16.60kN[/latex]

Now go on with the solution for [latex]F_e[/latex].

The correct answers are

[latex]Fe= W-F_b \quad\quad\quad\quad\quad F_e =F_b -W[/latex]

[latex]=10.86kN-16.60kN \quad\quad\quad\quad\quad =16.60kN-10.86kN[/latex]

[latex]=-5.74kN \quad\quad\quad\quad\quad = + 15.4kN[/latex]

Notice that both solutions yield the same numerical value, but they have opposite signs. The negative sign for the solution on the left means that the assumed direction for [latex]F_e[/latex] in Fig. 5.4(a) was wrong.

Therefore, both approaches give the same result.

Result Part (b) The required external force is a downward force of 5.74 kN.

How could you have reasoned from the start that a downward force would be required?

Items c and d of Step 4 of the procedure suggest that the specific weight of the cube and the fluid be compared. In this case we have the following results:

For the bronze cube [latex]\gamma_B = 86.9{kN}/{m^3}[/latex]

For the fluid (mercury) [latex]\gamma_m = \left(13.54\right)\left( 9.81{kN}/{m^3}\right) [/latex]

[latex]=132.8{kN}/{m^3}[/latex]

Comment Because the specific weight of the cube is less than that of the mercury, it would tend to float without an external force. Therefore, a downward force, as pictured in Fig. 5.4(b), would be required to hold it in equilibrium under the surface of the mercury.

This example problem is concluded.

Question 5.1: A cube 0.50 m on a side is made of bronze having a specific ...

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