Question

е Maps Translate News P Omin FIGURE 1 1) FIGURE 1 shows a uniform ladder PQ of weight 240 N leans on a smooth wall and resting on a rough floor with a minimum inclination angle Omin. The coefficient of friction between the ladder and floor is 0.25. (i) With the aid of a force diagram, calculate the forces acting on the ladder at point P and point Q. (ii) Determine the value of Omin- [15 marks]​

Answer 1

Answer:

Correct option is

C

μ=1/3

Let mass of ladder be m. consider free body diagram of ladder.

For verticle equilibrium -

N

1

+μN

2

=mg __(1)

For Horizontal equilibrium

N

2

=μN

1

__(2)

Torqul about point A

mg(2.5cosθ)=N

2

(4)+(μN

2

)

3

mg(2.5×

5

3

)=(4+3μ)N

2

(cosθ=

5

3

)

mg(

2

3

)=(4+3μ)N

2

___(3)

From (1) and (2) N

2

[

μ

1+μ

2

]=mg⇒N

2

=

1+μ

2

μmg

Hence mg(

2

3

)=(4+3μ)[

1+μ

2

μmg

]

(3+3μ

2

)=8μ+6μ

2

3μ

2

+6μ−3=0

⇒μ=

6

−8±

64+36

=

6

−8±10

μ=

3

1

Answer 2

$63^0$ is the required angle against the ground.

Explanation:

• From Equilibrium of forces;

$\Sigma F_x = 0\ in\ x\ axis\\\\\Sigma F_x = F_f - N_w_a_l_l\\\\\implies N_w_a_l_l = F_f\\\\\Sigma F_y = N_g_r_o_u_n_d - W = 0 \ in \ y \ axis\\\\\implies N_g_r =W = 240 N$

• From Equilibrium of moments;

$\frac{1}{2} W L \ cos(\theta) - N_w_a_l_l L \ sin (\theta) =0\\\\\implies tan (\theta) = \frac{W}{2N_w_a_l_l} = \frac{W}{2F_f}\\ \\Friction (F_f) = \mu N_g_r\\\\tan(\theta) = \frac{W}{2N_w_a_l_l} = \frac {W}{2\mu W} = \frac{1}{2\mu}\\\\\implies \theta = tan^-^1 (\frac{1}{2\mu}) = 63^0$