AssignmentXI-solution
1. A block of mass ‘m’ sits on an incline shown below. The incline is moving with an acceleration of
3m/s2.
FN
Ffict cos37°
f
37°
Ffict
Fg sin37°
a=3m/s 2
Ffict sin37°
37°
Fg cos37°
Fg
37°
i) What must be the frictional force between the block and incline if the block is not to slide along the
incline when the incline is accelerating to the right at 3m/s2? Ans) f =3.5m(N)
Sol) Our system is only enclosed on the block of the ramp. Ffict is the fictitious force being ‘applied’ to the
block due to the acceleration of the ramp.
When the net force on horizontal component zero, then block will not slide down.
ΣFx=Fg sin37° – f – Ffict cos37°= 0 and ΣFy=FN – Ffict sin37° – Fg cos37°= 0
Here, Ffict=ma, where ‘m’ is the mass of the block and ‘a’ is the acceleration of the ramp.
So in the x-component, fs =mgsin37°– ma cos37°=m(9.8×sin37°–3×cos37°)=3.5m(N)
ii) What is the least value of µs can have for the above to happen? Ans) μs=0.36
Sol) fs=μsFN, and from y-component FN = Ffict sin37° + Fg cos37°, so
μs=
3.5𝑚
=0.36
𝑚(3𝑠𝑖𝑛37°+9.8𝑐𝑜𝑠37°)
2. Find the acceleration of the cart that is required to prevent block B from falling. The coefficient of
static friction between the block and the cart is µs.
(hint: the horizontal forces acting on the block will be Ffict=max, where ‘m’ is the mass of the block and
‘a’ is the acceleration of the cart and the normal force FN)
system!!(any force from outside the system is treated as an external force)
ax
Y
f
Ffict
FN
X
Fg
Answer: c)
Sol) Our system(our focus of observation) is the block in front of the cart. If the cart accelerates, the
block will ‘feel’ a force(Ffict) pushing it against the front wall of the cart. This force is ‘real’ as far as the
block is concerned and we(=observer) are moving with the cart observing only inside the system.
ΣFx=FN – Ffict=0 , ΣFy=fs – mg=0
Here, Ffict=max, where ‘m’ is the mass of the cart and ‘ax’is the acceleration of the cart.
So from X-component; max = FN, and Y-component; f =mg
Since fs =µsFN, µsFN=mg, and since max = FN, then µsmax =mg. So ax = g/µs
3. A rectangular block of mass m sits on top of another similar block, which in turn sits on a flat table.
The maximum possible frictional force of one block on the other is f max= 2m(N).
Y
FN
system!
m
Ffict
f
X
Fg
i) What is the largest possible acceleration which can be given the lower block without the upper block
sliding off? Ans) 2m/s2
Sol) ΣFx= f – Ffict=0 , ΣFy=FN – mg=0
Since Ffict=ma, from X-component, f = ma, where ‘m’ the mass of the top block, and ‘a’ is the
acceleration of the bottom block.
And since the frictional force is given as f=2m,
2m=ma
a=2m/s2
ii) What is the coefficient of friction between the two blocks? Ans) 0.204
Sol) f =μsFN =μsmg, so μ=
𝑓
𝑚𝑔
=
2𝑚
𝑚𝑔
=
2
𝑔
=0.204