Question

A uniform ladder weighing 330 Newton is leaning against a wall. The ladder slips when the angle with the floor is 50 degrees. Assuming the coefficient of the static friction at the wall and the floor are the same, obtain a value for static friction.

Answer 1

Answer on Question #45180-Physics-Mechanics-Kinematics-Dynamics

A uniform ladder weighing 330 Newton is leaning against a wall. The ladder slips when the angle with the floor is 50 degrees. Assuming the coefficient of the static friction at the wall and the floor are the same; obtain a value for static friction.

Solution

The sum of angular momentum across point B is zero:

$$W \cdot \frac {L}{2} \cdot \cos 5 0 + F _ {A} L \sin 5 0 - N _ {A} L \cos 5 0 = 0.$$ $$F _ {A} = \mu N _ {A}, F _ {B} = \mu N _ {B}.$$ $$\sum F _ {x} = 0 \colon F _ {A} - N _ {B} = 0 \rightarrow F _ {A} = N _ {B}.$$ $$\sum F _ {y} = 0 \colon N _ {A} + F _ {B} - W = 0 \rightarrow W = N _ {A} + F _ {B}.$$

So

$$\left(N _ {A} + F _ {B}\right) \cdot \frac {1}{2} \cdot \cos 5 0 + F _ {A} \sin 5 0 - N _ {A} \cos 5 0 = 0.$$

The sum of angular momentum across point A is zero:

$$W \cdot \frac {L}{2} \cdot \cos 5 0 - N _ {B} L \sin 5 0 - F _ {B} L \cos 5 0 = 0.$$ $$\left(N _ {A} + F _ {B}\right) \cdot \frac {1}{2} \cdot \cos 5 0 - N _ {B} \sin 5 0 - F _ {B} \cos 5 0 = 0.$$

We can subtract equations across two points:

$$\left(F _ {A} + N _ {B}\right) \sin 5 0 = \left(N _ {A} - F _ {B}\right) \cos 5 0.$$

But $N_B = F_A$ ; $F_B = \mu N_B = \mu F_A$ . So

$$\tan 5 0 = \frac {N _ {A} - F _ {B}}{F _ {A} + N _ {B}} = \frac {N _ {A} - \mu F _ {A}}{2 F _ {A}} = \frac {\frac {F _ {A}}{\mu} - \mu F _ {A}}{2 F _ {A}} = \frac {\frac {1}{\mu} - \mu}{2}.$$ $$\mu^ {2} + 2 \mu \tan 5 0 - 1 = 0.$$ $$\mu = \frac {- 2 \tan 5 0 \pm \sqrt {4 (\tan^ {2} 5 0 + 1)}}{2}.$$

The positive root is physically possible. Therefore,

$$\mu = - \tan 5 0 + \sqrt {(\tan^ {2} 5 0 + 1)} = 0. 3 6.$$

Answer: 0.36.

http://www.AssignmentExpert.com/