Question 273: Laws of Motion

Question

A block of mass m
1
= 1 kg another mass m
2
= 2 kg, are placed together (see figure) on an inclined plane with angle of inclination $$	heta$$. Various values of $$	heta$$ are given in List I. The coefficient of friction between the block m
1
and the plane is always zero. The coefficient of static and dynamic friction between the block m
2
and the plane are equal to $$\mu$$ = 0.3. In List II expressions for the friction on the block m
2
are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g.
[Useful information : tan (5.5$$^\circ$$) $$\approx$$ 0.1; tan (11.5$$^\circ$$) $$\approx$$ 0.2; tan (16.5$$^\circ$$) $$\approx$$ 0.3]
List I
List II
P.
$$	heta  = 5^\circ $$
1.
$${m_2}g\sin 	heta $$
Q.
$$	heta  = 10^\circ $$
2.
$$({m_1} + {m_2})g\sin 	heta $$
R.
$$	heta  = 15^\circ $$
3.
$$\mu {m_2}g\cos 	heta $$
S.
$$	heta  = 20^\circ $$
4.
$$\mu ({m_1} + {m_2})g\cos 	heta $$

Accepted Answer

The forces on the block of mass m
1
are its weight m
1
g, normal reaction from the inclined plane N
1
, and the reaction from the second block R.
Similarly, forces on the block of mass m
2
are m
2
g, N
2
, R, and the frictional force f. If $$	heta$$ is slowly increased, f starts increasing and attains its maximum value f = $$\mu$$N
2
at $$	heta$$ = $$	heta$$
r
(angle of repose). The blocks are stationary if $$	heta$$ $$\le$$ $$	heta$$
r
otherwise they are moving. Consider the limiting case, $$	heta$$ = $$	heta$$
r
, when the blocks are at rest and
$$f = \mu {N_2}$$ ...... (1)
Apply Newton's second law to m
1
,
$$R = {m_1}g\sin 	heta $$ ....... (2)
$${N_1} = R = {m_2}g\cos 	heta $$ ....... (3)
and to m
2
,
$$f = R + {m_2}g\sin 	heta $$ ...... (4)
$${N_2} = {m_1}g\cos 	heta $$ ...... (5)
Eliminate R, N
2
and f from equations (1)-(5) to get
$${	heta _r} = {	an ^{ - 1}}\left( {{{\mu {m_2}} \over {{m_1} + {m_2}}}} ight) = {	an ^{ - 1}}\left( {{{0.3 	imes 2} \over {1 + 2}}} ight)$$
$$ = {	an ^{ - 1}}(0.2) = 11.5^\circ $$
Thus, for $$	heta$$ = 5$$^\circ$$ and $$	heta$$  = 10$$^\circ$$, blocks are at rest with frictional force
$$f = R + {m_2}g\sin 	heta  = ({m_1} + {m_2})g\sin 	heta $$
For $$	heta$$ = 15$$^\circ$$ and $$	heta$$ = 20$$^\circ$$, blocks are moving with frictional force
$$f = \mu {N_2} = \mu {m_2}g\cos 	heta $$, (limiting value).

Answer

P-1, Q-1, R-1, S-3

Answer

P-2, Q-2, R-2, S-3

Answer

P-2, Q-2, R-2, S-4

Answer

P-2, Q-2, R-3, S-3

Laws of Motion

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