JEE Physics Laws Of Motion Notes Set 05

Force and the Four Basic Interactions

A force is a push or a pull. It can change the speed of a body, change its direction of motion, change both, or merely deform it. Force is a vector, with SI unit newton, \( 1\ \text{N}=1\ \text{kg m s}^{-2} \), dimensions \( [MLT^{-2}] \), and \( 1\ \text{N}=10^5\ \text{dyne} \). A kilogram-force is the pull of the earth on 1 kg, so \( 1\ \text{kgf}=g\ \text{N} \).
Every force met in mechanics is a form of one of four basic interactions. Gravitation is the weakest, always attractive and of unlimited range; near the surface it appears as the weight \( mg \). The electromagnetic force can attract or repel, has long range and depends on the medium; tension, normal reaction, friction, muscular push and the spring force are all its macroscopic faces. The strong nuclear force binds nucleons together and acts only over about a femtometre. The weak force acts between elementary particles and is responsible for beta decay. For two protons 1 fm apart the relative strengths are roughly \( 1:10^{-2}:10^{-7}:10^{-38} \) for nuclear : electromagnetic : weak : gravitational.

Field Forces and Contact Forces

A field force acts across empty space through a field (gravitational and electric forces). A contact force appears only when bodies touch, and is split into two perpendicular parts: the normal force, which is the component perpendicular to the surfaces and always pushes, and friction, the tangential component that opposes relative sliding.
The normal force is a dependent force: it exists only as much as the surfaces are actually pressed together. Two blocks that merely touch on a smooth table, with nothing pushing them together, exert zero normal force on each other. It acts perpendicular to the common tangent at the point of contact, which for a rod resting on an edge means perpendicular to the rod, and for a point touching a surface means perpendicular to the surface.
Tension is the pulling force in a string. In a light string with no load between two points, the tension is the same everywhere; if a part of the string has mass or is accelerating, the tension changes along its length. To find it at a point, cut the string there and write Newton's second law for the piece on one side.

Newton's Third Law

Forces always occur in pairs: if A pulls or pushes B with force \( \vec F_{AB} \), B pushes or pulls A with \( \vec F_{BA}=-\vec F_{AB} \). Four points matter. The two forces act at the same instant, so neither is the cause of the other. They act on different bodies, so they can never cancel in the equation of one body. They are of the same kind (both gravitational, both normal, and so on). For a system made of both bodies, the pair is internal and the pair sums to zero, which is why internal forces drop out of the second law for a system.
A common slip: the weight of a block on a table and the normal force from the table are not a third-law pair, although they are equal in magnitude when the block is at rest. The partner of the weight is the block's pull on the earth; the partner of the normal force is the block's push on the table.

System, Internal and External Forces, Free Body Diagram

A system is any collection of bodies you decide to treat together. Forces between its members are internal; forces from outside it are external. A free body diagram (FBD) shows one body or sub-system isolated from its surroundings with only the external forces on it.
Recipe for an FBD: 1. Choose the body or system and imagine a boundary around it. 2. Draw the non-contact force first, normally the weight \( mg \) downwards. 3. Go around the boundary and draw every contact force where something touches it: normal reaction, friction, tension, applied push or pull. 4. Never draw internal forces, forces the body exerts on others, or the resultant itself. 5. Put in a coordinate system and resolve forces into components.

The Three Laws of Motion

First law: a body stays at rest or moves with constant velocity along a straight line unless a net external force acts on it. It also defines inertial frames, the frames in which a force-free body shows no acceleration. A frame at rest or moving with constant velocity is inertial; one that accelerates is not.
Second law: the net external force equals the rate of change of linear momentum, \( \vec F=\frac{d\vec p}{dt} \), which for constant mass is \( \vec F=m\vec a \). It is a vector law that works component by component: \( F_x=ma_x,\ F_y=ma_y,\ F_z=ma_z \). A force at an angle to the velocity changes only the velocity component along the force; the component perpendicular to the force is untouched. The law holds for a particle, F means the net external force, and it is local: the acceleration now depends on the force now, not on the history.
Third law: as above. Using the second law for two interacting bodies with no outside force gives \( \frac{d}{dt}(\vec p_1+\vec p_2)=0 \), so the third law and conservation of momentum are two statements of the same fact.

Solving Problems: Equilibrium and Acceleration

Procedure for equilibrium: sketch, isolate one body, draw its FBD, resolve forces along convenient axes, and write \( \sum F_x=0,\ \sum F_y=0 \). If two equations contain more than two unknowns, pick another body and repeat. Procedure for accelerating bodies is the same with \( \sum F_x=ma_x,\ \sum F_y=ma_y \). For a body connected to others, count unknowns against equations; when the unknowns outnumber the equations, a constraint relation is missing.
Handy results: two blocks of masses \( m_1,m_2 \) in contact on a smooth table pushed by a force \( F \) on \( m_1 \) share the acceleration \( a=\frac{F}{m_1+m_2} \) and the contact force is \( N=\frac{m_2F}{m_1+m_2} \). A uniform rod of mass \( M \) and length \( L \) pulled by \( F \) at one end on a smooth floor has tension \( T=F\left(1-\frac xL\right) \) at distance \( x \) from the pulled end. A rope of mass \( m \) accelerating upward at \( a \) has tension that grows linearly from the free end to the pulling end.

Weighing Machine, Lift Problems and Spring Balance

A weighing machine reads the force the object exerts on its upper surface, which equals the normal reaction \( N \). A spring balance reads the force at its hook, which equals the tension of the spring. Neither reads weight as such. For a man of mass \( m \) in a lift with upward acceleration \( a \): \( N-mg=ma \), so \( N=m(g+a) \). Upward acceleration makes the reading larger, downward acceleration \( N=m(g-a) \) smaller, and zero in free fall. Only the acceleration matters, not the velocity: a lift moving upward at constant speed gives the true weight, and so does one moving downward at constant speed.
To measure acceleration of an object at rest relative to a lift, compare the reading with \( mg \): a reading above \( mg \) means acceleration upward (the lift is speeding up while rising or slowing down while descending), and below \( mg \) means acceleration downward.

Spring Force

A spring resists any change of length: \( F=-kx \) where \( x \) is the change from the natural length and \( k \) is the stiffness in \( \text{N m}^{-1} \). The force is zero at natural length; it is the change of length that counts, not the position of the ends. If both ends of a spring shift by the same amount in the same direction, nothing is stretched. If one end moves 0.75 m towards the other and the other end moves 0.25 m towards it, the spring is compressed by 1 m and the force is \( k\times1 \).
The spring force cannot change instantly, because the length cannot change instantly. When a string holding a spring-and-block arrangement is cut, the spring force stays the same for that first instant while the string tension vanishes; the accelerations just after the cut follow from the new FBDs. A massless spring or spring balance in a line carries the same force at both ends, so its reading is that force, not the sum of the two end pulls.

String Constraint

An inextensible string that stays taut fixes a relation between the motions of the bodies it joins. Method: 1. Name the bodies and strings. 2. Assign a position variable to each body measured from a fixed point such as a pulley centre. 3. Write the total length of each string as a sum of its straight segments, equal to a constant. 4. Differentiate once for velocities, twice for accelerations, giving a sign \( + \) to a segment that lengthens and \( - \) to one that shortens.
Simple cases: for two masses over a fixed pulley, \( x_1+x_2=\text{const} \) gives \( a_1=a_2 \) in magnitude. For a movable pulley carried by a string whose ends move at velocities \( v_1 \) and \( v_2 \) (both measured the same way), the pulley moves with \( v_p=\frac{v_1+v_2}{2} \) and likewise for accelerations. If the string passes round a pulley and one end is pulled at speed \( v \) while the load hangs from a point at horizontal distance \( x \) from a pulley at height \( b \) above it, then \( \sqrt{x^2+b^2}+y=\text{const} \) gives the load velocity \( \frac{dy}{dt}=\frac{x}{\sqrt{x^2+b^2}}\frac{dx}{dt} \).
A quick rule that avoids calculus: the velocity components of two points joined by a taut string, taken along the string, are equal. The same rule applies to the ends of a rigid rod.

Wedge Constraint

When two rigid bodies stay in contact, the components of their velocities perpendicular to the contact surface are equal; otherwise they would separate or interpenetrate. Apply this at each contact. A rod sliding vertically at speed \( u \) on a wedge of face angle \( \theta \) that moves horizontally at speed \( v \) gives \( u\cos\theta=v\sin\theta \) in the geometry where the face makes angle \( \theta \) with the horizontal, so \( u=v\tan\theta \). A rigid rod of length \( l \) with ends on two perpendicular walls has end velocities with equal components along the rod: \( u\sin\theta=v\cos\theta \) so \( v=u\tan\theta \), and its midpoint traces a circle of radius \( l/2 \) about the corner.

Newton's Second Law for a System

For a set of bodies, \( \vec F_{ext}=m_1\vec a_1+m_2\vec a_2+m_3\vec a_3+\dots \), where internal forces cancel in pairs. This lets you skip internal forces, especially when the accelerations differ. For example, a block of mass \( m \) sliding down a smooth wedge of mass \( m \) free on a horizontal floor: no horizontal external force acts, so with wedge acceleration \( a \) and block acceleration \( b \) relative to the wedge along the slope, \( 0=ma+m(a-b\cos\theta) \). Combined with the block's own equation along the slope, \( mg\sin\theta=m(b-a\cos\theta) \), this gives \( a=\frac{g\sin\theta\cos\theta}{1+\sin^2\theta} \) and \( b=\frac{2g\sin\theta}{1+\sin^2\theta} \).

Non-Inertial Frames and Pseudo Force

In a frame accelerating with \( \vec a_{f} \), Newton's second law keeps its form if a pseudo force \( \vec F_{ps}=-m\vec a_{f} \) is added to the real forces: \( \vec F_{real}+\vec F_{ps}=m\vec a_{rel} \). The pseudo force points opposite to the acceleration of the frame (not opposite to its velocity), it has no third-law partner, and it needs no source body. A frame moving at constant velocity needs no pseudo force.
Lift and pendulum: a bob hangs from the ceiling of a lift accelerating upward at \( a \). An observer on the ground says \( T-mg=ma \). An observer inside sees the bob at rest, includes a pseudo force \( ma \) downward and gets \( T=mg+ma \) too. Both agree, which is the point.
Block on a wedge: for the block to stay at rest on a smooth wedge of angle \( \theta \) when a horizontal force \( F \) pushes the wedge, the common acceleration must satisfy \( N\cos\theta=mg,\ N\sin\theta=ma \), so \( a=g\tan\theta \) and \( F=(M+m)g\tan\theta \). Working in the wedge frame with a pseudo force \( ma \) gives the same equations.

Worked Examples

Example 1. A 10 kg block hangs from a string. The block is in equilibrium: \( T-10g=0 \), so \( T=100 \) N for \( g=10 \) m/s².

Example 2. A 10 kg block hangs from a knot A. A horizontal string pulls A towards a wall and a second string joins A to the ceiling at 30° to the vertical. Vertical balance at A: \( T_2\cos30^\circ=100 \), so \( T_2=\frac{200}{\sqrt3} \) N. Horizontal balance: \( T_1=T_2\sin30^\circ=\frac{100}{\sqrt3} \) N.

Example 3. A 5 kg block carries a 2 kg rope below it, and a 3 kg block hangs at the rope's lower end. An upward force \( F_0 \) accelerates everything upward at 2 m/s². Taking the whole system, \( F_0-100=10\times2 \), so \( F_0=120 \) N. The rope alone needs net force \( 2\times2=4 \) N. For the tension at the rope's midpoint, take the 3 kg block plus the lower 1 kg of rope: \( T-4g=4\times2 \), so \( T=48 \) N.

Example 4. Two masses on a light string over a smooth fixed pulley, \( m_1>m_2 \). The constraint gives equal acceleration magnitudes; the two equations \( m_1g-T=m_1a \) and \( T-m_2g=m_2a \) give \( a=\frac{m_1-m_2}{m_1+m_2}g \) and \( T=\frac{2m_1m_2}{m_1+m_2}g \). For 3 kg and 2 kg: \( a=2 \) m/s² and \( T=24 \) N.

Example 5. A 60 kg painter stands on a 15 kg platform and pulls the rope over an overhead pulley so that the rope tension equals his pull \( T \) and two rope segments support the pair. Then \( 2T-(75)g=75a \). With \( T=400 \) N: \( a=\frac{800-750}{75}\approx0.67 \) m/s². For an acceleration of 1 m/s² he must pull \( T=\tfrac12\times75\times11=412.5 \) N, and for constant speed \( T=\tfrac12\times75\times10=375 \) N.

Example 6. Two equal blocks A and B, each of mass \( m \), are joined by a light spring, and A hangs from the ceiling by a string, with B below the spring. In equilibrium the string pulls up with \( 2mg \) and the spring pulls B up with \( mg \). The instant the string is cut, the spring force \( mg \) is unchanged, so B has zero net force and \( a_B=0 \), while A feels its weight \( mg \) plus the spring pull \( mg \) downwards and \( a_A=2g \) downwards.
Variation: the spring hangs from the ceiling holding A, and a string joins A to B below. In equilibrium the spring force is \( 2mg \) and the string tension is \( mg \). When the string is cut, the spring still pulls A up with \( 2mg \): \( a_A=\frac{2mg-mg}{m}=g \) upwards, and B falls freely with \( a_B=g \) downwards.