The document Laws of Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev is a part of the JEE Course DC Pandey Solutions for JEE Physics.

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**Introductory Exercise 5.1**

**Q.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 cosθ = f

w + N sinθ = 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.

**Q.2. Two spheres A and B are placed between two vertical walls as shown in figure. Friction is absent everywhere. Draw the free body diagrams of both the spheres.**

Sol.

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_{A}= Weight of sphere A.

N' cos θ = R

N' sin θ + w_{A}= N

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_{B}= Weight of sphere B,

R', N(R) and w_{B}pass through point O, the centre sphere B.

**Q.3. A point A on a sphere of weight W rests in contact with a smooth vertical wall and is supported by a string joining a point B on the sphere to a point C on the wall. Draw free body diagram of the sphere.**

Sol:

N = Normal force on sphere by wall,

w = Weight of sphere,

T = Tension in string.

**Q.4. Write down the components of four forces along ox and oy directions as shown in Fig. 5.33.**

- F
_{1}_{x}= 2√3 N - F
_{2}x = -2 N - F
_{3}x = 0 - F
_{4}x = 4 N - F
_{1}_{y}= 2N - F
_{2}_{y}= 2√3 N - F
_{3}_{y}= -6 N - F
_{4}_{y}= 0

Component of

along x-axis : 4 cos 30° = 2√3 N

along y-axis : 4 sin 30° = 2 N

Component of

along x-axis : 4 cos 120° = - 2 N

along y-axis : 4 sin 120° = 2√3 N

Component of

along x-axis : 6 cos 270° = 0 N

along y-axis : 6 sin 270° = - 6 N

Component of

along x-axis : 4 cos 0° = 4 N

along y-axis : 4 sin 0° = 0 N

**Q.5. A uniform rod AB of weight W is hinged to a fixed point at A It is held in the horizontal position by a string, one end of which is attached to B as shown in Fig. 5.34. Find in terms of W, the tension in the string.**

Sol:Taking moment about point A

AB = 1

(T sin 30) 1 = w 1/2

⇒ T = w

**Q.6. In Question 3 of the same exercise, the radius of the sphere is a. The length of the string is also a. Find the tension in the string.**

Sol.See figure (answer to question no. 3)

T cos 30° = w

or

or

**Q.7. Find the values of the unknown forces if the given set of forces shown in Fig. 5.35 are in equilibrium.**

Sol.

F = 10.16 newton, A/= 2.4 newton

R cos 30° + 3 = f cos 60°

i.e.,

or

R√3 + 6 = f …(i)

and R sin 30° + f sin 60° = 10

i.e.,

or R + f√3 = 20 …(ii)

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

R + ( R√3 + 6)√3 = 20

4 R + 6√3 = 20

⇒

∴ f = (2.4) √3 + 6

= 10.16 N

**Q.8. Two beads of equal masses m are attached by a string of length √2a and are free to move in a smooth circular ring lying in a vertical plane as shown in Fig. 5.36. Here, a is the radius of the ring. Find the tension and acceleration of B just after the beads are released to move.**

Sol.

At point B (instantaneous vertical acceleration only)

∴ mg - T sin 45° = ma …(i)

At point A (instantaneous horizontal acceleration only)

∴ T cos 45° = ma …(ii)

Combining Eqs. (i) and (ii)

mg - ma = ma

⇒

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