Question

Mass of block is 8 kg. The minimal force required to move the block up if the coffiction of friction µ = 0.3. Take g = 10 m.s-2

Answer 1

Answer:

hey buddy ..

plzz refer to attachment..

hope it helps..

Answer 2

Answer:

67.3 N

Explanation:

An diagram is attached for better understanding.

  Here,

      Normal reaction is equal to mg.cos37°

So,

      Friction on block is μ . mg. cos37°

                   ⇒ 0.3 x mg x cos37°

Also, one of the component of weight is acted with the friction.

 Total force acted opposite the force for lifting is ( 0.3 mg . cos37° ) + ( mg sin37° )

Minimum force required = ( 0.3 mg cos37° ) + ( mg sin37° )

        ⇒ mg( 0.3 cos37° + sin37° )

You can use the approx values of cos37° and sin37°, here, I am directly writing 0.3 cos37° + sin37° as 0.8414      

⇒ 8 x 10( 0.8414 ) N

⇒ 8 x 8.414 N

⇒ 67.3 N ( approx )

Min. force required is 67.3 N