Question

Laws of Motion 149
4.
If block is at verge of motion, then select the
correct option (where u is the coefficient of friction)
[NCERT Pg. 101]
F
m
(1) Fumg
(2) F=
sine
umg
sin 0+ucose
(4) F = u mg
(3) F = mg​

Answer 1

Answer:

(2)

Explanation:

please see picture for answer

Answer 2

Given:

A block of mass m

A force F being applied on the block at an angle θ from the vertical

To Find:

The value of F

Solution:

The correct option is (2) F = μmg/(sinθ + μcosθ)

The vertical component of F = Fcosθ

The horizontal component of F = Fsinθ

Since the block is on the verge of motion,

⇒ The net force on the block should be zero

Balancing Vertical Forces:

Weight of the block = Normal Reaction + The vertical component of F

or mg = N + Fcosθ

or N = mg - Fcosθ    - (1)

Balancing Horizontal forces:

Frictional Force = The horizontal component of F

or μN = Fsinθ

Substituting the value of N from (1),

Fsinθ = μ(mg - Fcosθ) = μmg - μFcosθ

or F (sin θ + μcosθ) = μmg

or F = μmg / (sinθ + μcosθ)