Test: Common forces in mechanics (June 6) - NEET MCQ

• The coefficient of friction between a given pair of substances is independent of the area of contact between them.
• μ > 1 for copper on cast iron.
• Rolling friction is much smaller than sliding friction. Irregularity of the surfaces in contact is not a main contributor of friction

Static friction is a self adjusting force.

In the machines, the flow of compressed and purified air lowers the friction.
Hence it acts as a lubrican

A car accelerates on a horizontal road due to the force exerted by the road on the car.

The acceleration of the body sliding down smooth inclined plane is
a = gsinθ …(i)
The acceleration of the same body sliding down a rough inclined plane is
a′ = gsinθ − μgcosθ
= g (sinθ−μcosθ) ...(ii)
As the body slides a distance d in each case,
∴  (∵ u = 0)
∴  (∵ t′ = 20ps,t=20s)  
  (Using (i) and (ii))

sinθ = p²sinθ − μp² cosθ; μp² cosθ
= p²sinθ − sinθ
Here, θ=30∘ ∴ μ = 

The minimum force required to start pushing a body up a rough inclined plane is F₁ = mgsinθ + μgcosθ  ......(i)
Minimum force needed to prevent the body from sliding down the inclined plane is F₂ = mgsinθ − μgcosθ     .....(ii)
Divide (i) by (ii), we get

= 3 (tanθ=2μ (given))

As acceleration of the box is due to static friction,
∴ ma = f_s ≤ μ_s N = μ_s mg
a ≤ μ_sg
∴ a_max  = μg = 0.2 × 10ms⁻²
= 2ms⁻²

Limiting friction, f = μmg = 0.6 × 1 × 9.8
= 5.88 N
Applied force, F = ma = 1 × 5 = 5N
As F < f, so force of friction = 5 N

Let F be the force applied by the finger on the block. N is normal reaction of the wall on the block. Mg is weight of block acting vertically downwards. f is the force of friction between the wall and the block which acts upwards as shown in the figure.

 

The block will not fall, if f = Mg
μN = Mg  (∵ f = μN)
μF = Mg (∵ N = F)  = F = Mg/μ

As the string is inextensible, and the pulley is smooth, therefore both the 4kg block and the 20kg trolley have the same acceleration a.
Let T be tension in the string.
The equation of motion of the block is
40 − T = 4a  …(i)
The equation of motion of the trolley is
T − f_k  = 20a
f_k = μ_k N

Here, μ_k =0.02, N = 20 × 10 = 200N

∴ T − 0.02 × 200 = 20a or
T − 4 = 20a …(ii)
Adding (i) and (ii), we get, 
 a = 36 / 24 m s⁻²  = 1.5 m s⁻²