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# Laws of Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

## JEE : Laws of Motion (Part - 1) - Physics, Solution by DC Pandey NEET Notes | EduRev

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,
wA = Weight of sphere A.
N' cos θ = R
N' sin θ + wA = 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,
wB = Weight of sphere B,
R', N(R) and wB 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.  Sol.

1. F1x = 2√3 N
2. F2x = -2 N
3. F3x = 0
4. F4x = 4 N
5. F1y = 2N
6. F2y = 2√3 N
7. F3y = -6 N
8. F4y = 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 Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

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