# Suppose the Friction Coefficient Between the Ground and the Ladder is 0⋅540. Find the Maximum Weight of a Mechanic Who Could Go up and Do the Work Form the Same Position of the Ladder. - Physics

Sum

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 N2 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.

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 42 | Page 198