Question

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answer 1

Answer:

Explanation:

Given

Weight of ladder $W=100\ N$

coefficient of friction $\mu =0.4$

Length of ladder $L=8\ m$

$\sum F_x=0$

$\sum F_y=0$

$\sum _M=0$

from diagram

$f_r=N_1$

$N_2=W$

and $f_r=\mu N_2$

$f_r=\mu W$

taking moment about lowest point of ladder of length L

$N_1\times L\sin \theta -W\cos \theta \times \frac{L}{2}=0$

$\tan \theta =\frac{W}{2N_1}$

and $N_1=\mu W$

thus $\tan \theta =\frac{1}{2\mu }$

$\tan \theta =\frac{1}{2\times 0.4}$

$\tan \theta =1.25$

$\theta=51.34^{\circ}$

                           

Answer 2

Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

$F_{y}=n-mg$

$N-mg=0$

$N=mg$

We need to calculate the normal force

Using Newtons law in horizontal direction

$F_{s}=f_{s}-P$

$f_{s}-P=0$

$f_{s}=P$

$P=\mu mg$

We need to calculate the minimum angle

Using torque about the point A then

$-mg\cos\theta\times AB+P\sin\theta\times AC=0$

Put the value into the formula

$mg\cos\theta\times(\dfrac{L}{2})=\mu mg\sin\theta\times L$

$\cos\theta\times\dfrac{1}{2}=\mu\sin\theta$

$\dfrac{1}{2}=\mu\times\tan\theta$

$\theta=\tan^{-1}(\dfrac{1}{2\times0.40})$

$\theta=51.34^{\circ}$

Hence, The minimum angle is 51.34°