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__Introductory Exercise 5.1__

**Ques 1: The diagram shows a rough plank resting on a cylinder with one end of the plank on rough ground. Neglect friction between plank and cylinder. Draw diagrams to show: (a) the forces acting on the plank, (b) the forces acting on the cylinder.Sol: **N = Normal force on cylinder by plank

R = Normal force on cylinder by ground,

f = Force of friction by ground by cylinder,

w = Weight of cylinder.

N cosÎ¸ = f

w + N sinÎ¸ = R,

N(R) = Reaction to N,

(Force acting on plank)

i.e.,normal force on plank by cylinder

R' = Normal force on plank by ground,

w = Weight of plank,

f' = frictional force on plank by ground.

Resultant of f' and R', N(R) and w pass through point O.

R = Normal force on sphere A by left wall,

N = Normal force on sphere A by ground,

N' = Normal force on sphere A by sphere B,

w

N' cos Î¸ = R

N' sin Î¸ + w

R' = Normal force on sphere B by right wall,

N ( R) = Reaction to N i.e. normal force on sphere B by sphere A,

w

R', N(R) and w

Sol:

w = Weight of sphere,

T = Tension in string.

Ans:

along x-axis : 4 cos 30Â° = 2âˆš3 N

along y-axis : 4 sin 30Â° = 2 N

Component of

along x-axis : 4 cos 120Â° = - 2 N

along y-axis : 4 sin 120Â° = 2âˆš3 N

Component of

along x-axis : 6 cos 270Â° = 0 N

along y-axis : 6 sin 270Â° = - 6 N

Component of

along x-axis : 4 cos 0Â° = 4 N

along y-axis : 4 sin 0Â° = 0 N

Ans:

AB = 1

(T sin 30) 1 = w 1/2

â‡’ T = w

Ans:

T cos 30Â° = w

or

or

Ans:

i.e.,

or

Râˆš3 + 6 = f â€¦(i)

and R sin 30Â° + f sin 60Â° = 10

i.e.,

or R + fâˆš3 = 20 â€¦(ii)

Substituting the value of f from Eq. (i) in Eq. (ii)

R + ( Râˆš3 + 6)âˆš3 = 20

4 R + 6âˆš3 = 20

â‡’

âˆ´ f = (2.4) âˆš3 + 6

= 10.16 N

Ans:

âˆ´ mg - T sin 45Â° = ma â€¦(i)

At point A (instantaneous horizontal acceleration only)

âˆ´ T cos 45Â° = ma â€¦(ii)

Combining Eqs. (i) and (ii)

mg - ma = ma

â‡’

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