What minimum force F is needed to lift the piano (mass M) using the pulley apparatus shown in Fig. 4-66? (b) Determine the tension in each section of rope: $F_{\mathbf{T}\mathbf{1}}$, $F_{\mathbf{T}2}$, $F_{\mathbf{T}3}$and $F_{\mathbf{T}4}$. Assume pulleys are massless and frictionless, and that ropes are massless.

Figure 4-66Problem 76

The net force acting on the body is equal to the product of the mass of the body multiplied by the acceleration of the body.

As the string is massless, the net force on the string must be zero according to Newton’s second law, and the tension throughout the length of the rope must be the same. Thus,

$F_{\mathrm{T}1}=F\dots \left(i\right)$

And

$F_{\mathrm{T}2}=F\dots \left(ii\right)$

For a limiting case, you can assume the piano to be in equilibrium.

$F_{\mathrm{T}4}=Mg\dots \left(iii\right)$

The FBD of the upper pulley can be drawn as:

Applying Newton’s second law, you get:

$F_{\mathrm{T}3}=F_{\mathrm{T}2}+F_{\mathrm{T}1}+F$

Using equations (i) and (ii), you get:

role="math" localid="1645682480034" $F_{\mathrm{T}3}=3F\dots \left(iv\right)$

The FBD of the lower pulley can be drawn as:

Applying Newton’s second law, you get:

$F_{\mathrm{T}2}+F_{\mathrm{T}1}=F_{\mathrm{T}4}$

Substituting the values from (i), (ii), and (iii), you get:

$2F=Mg$

From the above equation,

$F=\frac{Mg}{2}$

Thus, the minimum value of force F required to lift the piano is $\frac{Mg}{2}$.

Using equation (i), you will obtain:

$F_{\mathrm{T}1}=\frac{Mg}{2}$

Using equation (ii), you will obtain:

$F_{\mathrm{T}2}=\frac{Mg}{2}$

Using equation (iv), you will obtain:

$F_{\mathrm{T}3}=\frac{3Mg}{2}$

And, from (iii),

$F_{\mathrm{T}4}=Mg$

Thus, the values of $F_{\mathrm{T}1}$, $F_{\mathrm{T}2}$, $F_{\mathrm{T}3}$, $F_{\mathrm{T}4}$and are $\frac{Mg}{2}$, $\frac{Mg}{2}$, $\frac{3Mg}{2}$, and Mg, respectively.