Question

A block of Mass 1 kg sits on an incline at an angle 37°
A) what must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at 3m/se
B) what is the least value of friction cofficient can have for this to happen?

Answer 1

Answer:

For not sliding frictional force should be

$\huge \pink{f \geqslant \frac{42}{5} }$

And Coefficient

$\huge \green{ \mu \geqslant \frac{42}{31} }$

Answer 2

Answer:

When taking,$g \: = 10m {s}^{ - 2}$

A) f = 3.6 N

B) n = 0.36

When taking,

$g = 9.8m {s}^{ - 2}$

A) f = 3.48 N

B) n = 0.36

Explanation:

Apply pseudo-force on the block in opposite direction of the acceleration on incline.

Write up all the forces and their components by making a free body diagram.

Take the equation that is parallel to the incline plane for the value of f.

$f + ma.cos37 = mg.sin37 \\ f = mg.sin37 - ma.cos37 \\ f = (1)(10)( \frac{3}{5} ) - (1)(3)( \frac{4}{5} ) \\ f = \frac{30}{5} - \frac{12}{5} = \frac{18}{5} = 3.6 \: newton$

The equation for friction coefficient is:

$f = n \times normal \: reaction(nr) \\ n = \frac{f}{nr}$

For value of normal reaction, take the equation perpendicular to the incline plane.

$nr = mg.cos37 + ma.sin37 \\ nr = (1)(10)( \frac{4}{5} ) + (1)(3)( \frac{3}{5} ) \\ nr = \frac{40}{5} + \frac{9}{5} = \frac{49}{5} = 9.8 \: newton$

Then value of n will be:

$n = \frac{f}{nr} = \frac{3.6}{9.8} = 0.36$