Suppose the friction coefficient between the ground and

the ladder of the previous problem is 0.540. Find the

maximum weight of a mechanic who could go up and do

the work from the same position of the ladder.

#### Solution

Let the maximum mass of a mechanic who could go up be m.

The system is in translation and rotational equilibrium; therefore, we have

\[N_2 = 16 g + mg ...........(1)\]

\[ N_1 = \mu N_2.........(2)\]

\[ R_1 \times 10\cos37^\circ = 16g \times 5\sin37^\circ - mg \times 8 \sin37^\circ .........(3)\]

\[\Rightarrow 8 N_1 = 48g + \frac{24}{5}mg\]

From eq. (2), we have

\[ N_2 = \frac{48 g + \frac{24}{5} mg}{8 \times 0 . 54}\]

Putting the value of *N*_{2} in eq.(1), we have

\[16g + mg = \frac{24 . 0g + 24mg}{5 \times 8 \times 0 . 54}\]

\[ \Rightarrow 16 + m = \frac{24 . 0 + 24m}{40 \times 0 . 54}\]

\[ \Rightarrow m = 44 kg\]

Therefore, weight of the mechanic, who can go up and do the work, should be less than 44 kg.