SATHEE: Laws of Motion 3 Question 7

Laws of Motion 3 Question 7

7. Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$ . The system is in equilibrium. Find direction and magnitude of frictional force acting at $A$ or $B$ .

(2005)

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Answer:

Correct Answer: 7. (a)

Solution:

A block of mass $m$ is placed on a surface with a vertical cross-section, then

$\tan θ = \frac{d y}{d x} \frac{d \frac{x^{3}}{6}}{d x} = \frac{x^{2}}{2}$

At limiting equilibrium, we get

$\begin{aligned} μ & = \tan θ \\ \Rightarrow & 0.5 & = \frac{x^{2}}{2} \\ \Rightarrow & x^{2} & = 1 \\ x & = \pm 1 \end{aligned}$

Now, putting the value of $x$ in $y = \frac{x^{3}}{6}$ , we get

$\begin{array}{ll} When x = 1 & When x = - 1 \\ y = \frac{(1)^{3}}{6} = \frac{1}{6} & y = \frac{(- 1)^{3}}{6} = \frac{- 1}{6} \end{array}$

So, the maximum height above the ground at which the block can be placed without slipping is $1 / 6 m$ .

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Laws of Motion 3 Question 7

7. Two identical ladders are arranged as shown in the figure. Mass of each ladder is M and length L. The system is in equilibrium. Find direction and magnitude of frictional force acting at A or B.

Answer:

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7. Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$ . The system is in equilibrium. Find direction and magnitude of frictional force acting at $A$ or $B$ .