Question

(b) Derive an expression for the maximum value of θ for which the block will not slide down the incline.

Answer 1

The value of θ depends on the coefficient of static friction and the maximum value is obtained when coefficient of static friction is 1.

For a block inclined to a plane, the block will be subjected to horizontal and vertical forces.

The vertical component of force on the block is calculated as;

$F_n = mgcos(\theta)$

The horizontal component of the force on the block is calculated;

$mgsin(\theta) - F_k = ma$

The block will not slide down the inclined plane when the velocity is zero.

$mgsin(\theta) -F_k = 0\\\\mgsin(\theta) = F_k$

The coefficient of friction is given as;

$\mu_s = \frac{F_k}{F_n}= \frac{mgsin(\theta)}{mgcos(\theta)} \\\\\mu_s = \frac{sin(\theta)}{cos(\theta)} \\\\\mu_s = tan(\theta)$

where;

• $\mu_s$ is the coefficient of static friction between 0 and 1

Thus, the value of θ depends on the coefficient of static friction and the maximum value is obtained when coefficient of static friction is 1.

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