Test: Common Forces in Mechanics (NCERT)

• 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

• Both static and kinetic friction are independent on the area of contact.
• Coefficient of static friction depends on the surfaces in contact. μ_k < μ_s

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

Friction forces are always parallel to the surfaces in contact, which in this case, are the wall and the cover of the book. This tells us that the friction force is either upwards or downwards. Because the tendency of the book is to fall due to gravity, the friction force must be in the upwards direction.

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

The various forces acting on the block are as shown in the figure. From figure

mgsinθ = f     ....(i)
​mgcosθ = N     ....(i)
Divide (i) by (ii), we get
tanθ =  or
θ = tan⁻¹(μ) 

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))

Here, m = 10 kg, g= 10 m s^-2, μ = 0.5, F = 100N
Force of friction
f = μN = μMg = (0.5)(10)(10) = 50N

Force  that produce acceleration
F' = F - f = 100 N - 50 N = 50 N
a = F'/ m = 50 N / 10 kg = 5 m s^-2

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

Here, frictional force, f = μR
= μmgcosθ = 0.7 × 2 × 9.8cos30^∘
= 0.7 × 2 × 9.8 × 0.866 = 11.9 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