Translational Equilibrium and Newton's Laws Notes

Newton's First Law and Translational Equilibrium

  • Newton's First Law of Motion:

    • If an object is subject to no net force, then the object will not change its state of motion (its direction and speed remain constant, meaning it will not accelerate).

    • Conversely, if an object is observed to not change its state of motion, it can be concluded that the net force acting on the object is zero (∑F=0\sum \mathbf{F} = 0).

  • Net Force Definition:

    • The net force Fnet\mathbf{F}_{\text{net}} is the vector sum of all external forces acting on an object:     Fnet=∑F\mathbf{F}_{\text{net}} = \sum \mathbf{F}

  • Condition for Translational Equilibrium:

    • If ∑F=0\sum \mathbf{F} = 0, the object is in translational equilibrium.

    • If an object is in translational equilibrium, one can directly infer that ∑F=0\sum \mathbf{F} = 0

Inertial Reference Frames and Free-Body Diagrams

  • Inertial Reference Frame Caveat:

    • Newton's First Law and the principles of translational equilibrium are strictly valid in an inertial reference frame.

    • An inertial reference frame is defined as a coordinate system where the basis vectors (such as i^\hat{i}, j^\hat{j}, k^\hat{k}) do not change over time.

  • Free-Body Diagrams (FBD):

    • A free-body diagram is a fundamental tool for physical analysis.

    • Procedure for constructing a free-body diagram:

    • Model the physical object as a single zero-dimensional point particle.

    • Identify, list, and draw all external forces exerted on the object as vectors originating from the point particle.

Multi-Dimensional Equilibrium Calculations

  • System Configuration with Three Ropes:

    • A 10 kg10\,\text{kg} mass is suspended in static translational equilibrium in the xzxz-plane supported by three ropes.

    • Mass m=10 kgm = 10\,\text{kg}, with gravitational force given by:     Fg=−mgk^=−(10 kg)(9.8 m/s2)k^=−98 Nk^\mathbf{F}_g = -m g \hat{k} = -(10\,\text{kg})(9.8\,\text{m/s}^2)\hat{k} = -98\,\text{N}\hat{k}

    • Rope 1: Pulls with a force magnitude of 80 N80\,\text{N} at an angle of 30∘30^\circ relative to i^\hat{i} (and 60∘60^\circ relative to k^\hat{k}):     F1=(80 Ncos⁡(30∘))i^+(80 Ncos⁡(60∘))k^=69.3 Ni^+40.0 Nk^\mathbf{F}_1 = (80\,\text{N} \cos(30^\circ))\hat{i} + (80\,\text{N} \cos(60^\circ))\hat{k} = 69.3\,\text{N}\hat{i} + 40.0\,\text{N}\hat{k}

    • Rope 2: Pulls with a force magnitude of 50 N50\,\text{N} at an angle of 135∘135^\circ relative to i^\hat{i} (which is 45∘45^\circ relative to −i^-\hat{i}) and 45∘45^\circ relative to k^\hat{k}:     F2=(50 Ncos⁡(135∘))i^+(50 Ncos⁡(45∘))k^=−35.4 Ni^+35.4 Nk^\mathbf{F}_2 = (50\,\text{N} \cos(135^\circ))\hat{i} + (50\,\text{N} \cos(45^\circ))\hat{k} = -35.4\,\text{N}\hat{i} + 35.4\,\text{N}\hat{k}

    • Rope 3: Located in the xzxz-plane with unknown force components F3=F3xi^+F3zk^\mathbf{F}_3 = F_{3x}\hat{i} + F_{3z}\hat{k}, making an angle θ\theta measured counterclockwise from i^\hat{i}.

  • Equilibrium Vector Component Equations:

    • Setting the net force vector to zero:     Fnet=F1+F2+F3+Fg=0\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + \mathbf{F}_g = 0

    • Separating into scalar component equations:

    • x-component equation:       69.3 N−35.4 N+F3x=069.3\,\text{N} - 35.4\,\text{N} + F_{3x} = 0       33.9 N+F3x=0  ⟹  F3x=−33.9 N33.9\,\text{N} + F_{3x} = 0 \implies F_{3x} = -33.9\,\text{N}

    • z-component equation:       40.0 N+35.4 N−98.0 N+F3z=040.0\,\text{N} + 35.4\,\text{N} - 98.0\,\text{N} + F_{3z} = 0       −22.6 N+F3z=0  ⟹  F3z=22.6 N-22.6\,\text{N} + F_{3z} = 0 \implies F_{3z} = 22.6\,\text{N}

  • Magnitude and Direction of Force in Rope 3:

    • Magnitude:     ∣F3∣=F3x2+F3y2+F3z2=(−33.9 N)2+02+(22.6 N)2=40.7 N|\mathbf{F}_3| = \sqrt{F_{3x}^2 + F_{3y}^2 + F_{3z}^2} = \sqrt{(-33.9\,\text{N})^2 + 0^2 + (22.6\,\text{N})^2} = 40.7\,\text{N}

    • Direction Angle θ\theta:     F3x=∣F3∣cos⁡(θ)F_{3x} = |\mathbf{F}_3| \cos(\theta)     −33.9 N=40.7 Ncos⁡(θ)-33.9\,\text{N} = 40.7\,\text{N} \cos(\theta)     cos⁡(θ)=−33.940.7=−0.8329  ⟹  θ=146∘\cos(\theta) = \frac{-33.9}{40.7} = -0.8329 \implies \theta = 146^\circ

  • Alternative Component Projection Method:

    • Direct projection using angle projections:     F3x=−∣F3∣cos⁡(90∘)−80 Ncos⁡(30∘)−50 Ncos⁡(135∘)=−33.9 NF_{3x} = -|\mathbf{F}_3| \cos(90^\circ) - 80\,\text{N} \cos(30^\circ) - 50\,\text{N} \cos(135^\circ) = -33.9\,\text{N}     F3z=−98 Ncos⁡(180∘)−80 Ncos⁡(60∘)−50 Ncos⁡(45∘)=22.6 NF_{3z} = -98\,\text{N} \cos(180^\circ) - 80\,\text{N} \cos(60^\circ) - 50\,\text{N} \cos(45^\circ) = 22.6\,\text{N}

Contact Forces: Normal Force and Friction

  • Definition of Contact Forces:

    • Contact forces are forces exerted between two macroscopic objects in physical touch.

  • Action-Reaction Principle:

    • Contact forces obey Newton's Third Law: When object A and object B interact, the force exerted by A on B is equal in magnitude and opposite in direction to the force exerted by B on A:     FA on B=−FB on A\mathbf{F}_{\text{A on B}} = -\mathbf{F}_{\text{B on A}}

  • Nature as Forces of Constraint:

    • Contact forces are forces of constraint; they adjust dynamically to whatever magnitude and direction are required to maintain spatial constraints (preventing solid objects from interpenetrating or passing through one another).

  • Ideal Surface Conventions:

    • An ideal surface is an impenetrable interface.

    • Normal Force (FN\mathbf{F}_N   or Fnormal\mathbf{F}_{\text{normal}}): The component of the surface force acting at an angle of 90∘90^\circ (perpendicular) to the boundary plane.

    • Friction Force (Ff\mathbf{F}_f   or Ffriction\mathbf{F}_{\text{friction}}): The component of the surface force acting along (parallel to) the boundary plane.

Inclined Plane Equilibrium Analysis

  • System Properties:

    • A mass m=5 kgm = 5\,\text{kg} remains stationary on a rough slope inclined at an angle θ=25∘\theta = 25^\circ above the horizontal.

    • An external pushing force T\mathbf{T} of magnitude T=20 NT = 20\,\text{N} acts parallel to the slope surface, directed downward along the incline.

  • Global Vector Method for Net Surface Force:

    • Gravitational force: Fg=−mgk^=−(5 kg)(9.8 m/s2)k^=−49 Nk^\mathbf{F}_g = -m g \hat{k} = -(5\,\text{kg})(9.8\,\text{m/s}^2)\hat{k} = -49\,\text{N}\hat{k}.

    • Applied pushing force T\mathbf{T} has an angle of 155∘155^\circ relative to i^\hat{i} and 115∘115^\circ relative to k^\hat{k}:     T=20 Ncos⁡(155∘)i^+20 Ncos⁡(115∘)k^=−18.1 Ni^−8.4 Nk^\mathbf{T} = 20\,\text{N} \cos(155^\circ)\hat{i} + 20\,\text{N} \cos(115^\circ)\hat{k} = -18.1\,\text{N}\hat{i} - 8.4\,\text{N}\hat{k}

    • Net equilibrium condition for total force exerted by the slope Fslope\mathbf{F}_{\text{slope}}:     Fnet=Fg+T+Fslope=0\mathbf{F}_{\text{net}} = \mathbf{F}_g + \mathbf{T} + \mathbf{F}_{\text{slope}} = 0     Fslope=−(Fg+T)=−(−18.1 Ni^−8.4 Nk^)−(−49 Nk^)=18.1 Ni^+57.4 Nk^\mathbf{F}_{\text{slope}} = -(\mathbf{F}_g + \mathbf{T}) = -(-18.1\,\text{N}\hat{i} - 8.4\,\text{N}\hat{k}) - (-49\,\text{N}\hat{k}) = 18.1\,\text{N}\hat{i} + 57.4\,\text{N}\hat{k}

    • Magnitude of the total force exerted by the slope:     ∣Fslope∣=(18.1 N)2+(57.4 N)2=60.2 N|\mathbf{F}_{\text{slope}}| = \sqrt{(18.1\,\text{N})^2 + (57.4\,\text{N})^2} = 60.2\,\text{N}

    • Alternative Vector Addition Method:

    • Angle between gravity Fg\mathbf{F}_g (49 N49\,\text{N} vertical) and pushing force T\mathbf{T} (20 N20\,\text{N} along incline) is 90∘−25∘=65∘90^\circ - 25^\circ = 65^\circ.

    • Using vector law of cosines:       ∣Fslope∣2=(20 N)2+(49 N)2+2(49 N)(20 N)cos⁡(65∘)|\mathbf{F}_{\text{slope}}|^2 = (20\,\text{N})^2 + (49\,\text{N})^2 + 2(49\,\text{N})(20\,\text{N}) \cos(65^\circ)       ∣Fslope∣2=400+2401+828.3=3629.3  ⟹  ∣Fslope∣=60.2 N|\mathbf{F}_{\text{slope}}|^2 = 400 + 2401 + 828.3 = 3629.3 \implies |\mathbf{F}_{\text{slope}}| = 60.2\,\text{N}

  • Decomposed Surface Force Components:

    • Normal Force Component (FNF_N   ):

    • Perpendicular to incline plane (90∘90^\circ):       FN=mgcos⁡(25∘)=49 Ncos⁡(25∘)=44.4 NF_N = m g \cos(25^\circ) = 49\,\text{N} \cos(25^\circ) = 44.4\,\text{N}

    • Friction Force Component (FfF_f   ):

    • Parallel to incline plane:       Ff=T+mgsin⁡(25∘)=20 N+49 Nsin⁡(25∘)=20 N+20.7 N=40.7 NF_f = T + m g \sin(25^\circ) = 20\,\text{N} + 49\,\text{N} \sin(25^\circ) = 20\,\text{N} + 20.7\,\text{N} = 40.7\,\text{N}

    • Total Surface Force Consistency Check:     ∣Fslope∣=FN2+Ff2=(44.4 N)2+(40.7 N)2=60.2 N|\mathbf{F}_{\text{slope}}| = \sqrt{F_N^2 + F_f^2} = \sqrt{(44.4\,\text{N})^2 + (40.7\,\text{N})^2} = 60.2\,\text{N}

Laws of Static and Kinetic Friction

  • Static Friction (fsf_s ):

    • When an applied tangential force TT is small, static friction holds the object motionless.

    • As TT increases, it reaches a threshold where sliding begins. The minimum applied force required to initiate motion increases if the mass mm increases.

    • Mathematical inequality governing static friction:     fs≤fs,max=μsFNf_s \le f_{s,\text{max}} = \mu_s F_N

    • μs\mu_s is the dimensionless coefficient of static friction, which depends exclusively on the specific materials in contact.

    • Static friction matches whatever value is necessary to prevent motion up to its maximum capacity fs,maxf_{s,\text{max}}.

  • Kinetic Friction (fkf_k ):

    • Once an object begins moving relative to the surface, the force transitions to kinetic friction.

    • Mathematical equality governing kinetic friction:     fk=μkFNf_k = \mu_k F_N

    • μk\mu_k is the dimensionless coefficient of kinetic friction.

    • The direction of the kinetic friction force is always strictly opposite to the direction of motion.

Critical Inclines and Pulling Mechanics

  • Static Equilibrium and Sliding on an Adjustable Incline:

    • System parameters: Mass m=4 kgm = 4\,\text{kg}, coefficient of static friction μs=0.3\mu_s = 0.3, coefficient of kinetic friction μk=0.2\mu_k = 0.2.

    • Normal force component: FN=mgcos⁡(θ)F_N = m g \cos(\theta).

    • Down-slope gravitational component: Fg,∥=mgsin⁡(θ)F_{g,\parallel} = m g \sin(\theta).

    • Static Equilibrium Limit:     fs=mgsin⁡(θ)≤μsmgcos⁡(θ)f_s = m g \sin(\theta) \le \mu_s m g \cos(\theta)     tan⁡(θ)≤μs\tan(\theta) \le \mu_s     tan⁡(θ)≤0.3  ⟹  θ≤arctan⁡(0.3)=16.7∘\tan(\theta) \le 0.3 \implies \theta \le \arctan(0.3) = 16.7^\circ

    • Range of angles for static equilibrium: 0∘≤θ≤16.7∘0^\circ \le \theta \le 16.7^\circ

    • Constant Speed Sliding Condition:

    • Motion at constant speed implies acceleration a=0a = 0, so net force remains zero (Fnet=0\mathbf{F}_{\text{net}} = 0).     fk=mgsin⁡(θ)=μkmgcos⁡(θ)f_k = m g \sin(\theta) = \mu_k m g \cos(\theta)     tan⁡(θ)=μk\tan(\theta) = \mu_k     tan⁡(θ)=0.2  ⟹  θ=arctan⁡(0.2)=11.3∘\tan(\theta) = 0.2 \implies \theta = \arctan(0.2) = 11.3^\circ

  • Pulling a Mass on a Rough Horizontal Surface:

    • System parameters: Mass m=20 kgm = 20\,\text{kg}, weight W=mg=(20 kg)(9.8 m/s2)=196 NW = m g = (20\,\text{kg})(9.8\,\text{m/s}^2) = 196\,\text{N}.

    • Rope pulls at an angle θ=20∘\theta = 20^\circ above the horizontal.

    • Coefficients of friction: μs=0.4\mu_s = 0.4, μk=0.3\mu_k = 0.3

    • Forces acting on mass:

    • Tension T=Tcos⁡(θ)i^+Tsin⁡(θ)k^\mathbf{T} = T \cos(\theta)\hat{i} + T \sin(\theta)\hat{k}

    • Gravity Fg=−mgk^=−196 Nk^\mathbf{F}_g = -m g \hat{k} = -196\,\text{N}\hat{k}

    • Normal force FN=FNk^\mathbf{F}_N = F_N \hat{k}

    • Friction force f=−fi^\mathbf{f} = -f \hat{i}

    • Equilibrium equations:

    • Horizontal: fs=Tcos⁡(θ)f_s = T \cos(\theta)

    • Vertical: FN=mg−Tsin⁡(θ)F_N = m g - T \sin(\theta)

    • Important Rule: Normal force FNF_N must always be determined from the vertical physical equilibrium requirement that the mass does not penetrate the surface.

  • Minimum Tension to Initiate Motion:

    • Setting static friction to its limit:     Tcos⁡(θ)≤μs(mg−Tsin⁡(θ))T \cos(\theta) \le \mu_s (m g - T \sin(\theta))     Tcos⁡(θ)+μsTsin⁡(θ)≤μsmgT \cos(\theta) + \mu_s T \sin(\theta) \le \mu_s m g     T≤μsmgcos⁡(θ)+μssin⁡(θ)T \le \frac{\mu_s m g}{\cos(\theta) + \mu_s \sin(\theta)}

    • Numerical evaluation with μs=0.4\mu_s = 0.4, θ=20∘\theta = 20^\circ, mg=196 Nm g = 196\,\text{N}:     μsmg=0.4×196 N=78.4 N\mu_s m g = 0.4 \times 196\,\text{N} = 78.4\,\text{N}     cos⁡(20∘)+0.4sin⁡(20∘)=0.9397+0.4(0.3420)=0.9397+0.1368=1.0765\cos(20^\circ) + 0.4 \sin(20^\circ) = 0.9397 + 0.4(0.3420) = 0.9397 + 0.1368 = 1.0765     Tthreshold=78.4 N1.0765=72.8 NT_{\text{threshold}} = \frac{78.4\,\text{N}}{1.0765} = 72.8\,\text{N}

    • Static equilibrium range: T≤72.8 NT \le 72.8\,\text{N}. Motion starts when tension exceeds 72.8 N72.8\,\text{N}.

  • Tension for Motion at Constant Speed:

    • Using kinetic friction μk=0.3\mu_k = 0.3:     T=μkmgcos⁡(θ)+μksin⁡(θ)T = \frac{\mu_k m g}{\cos(\theta) + \mu_k \sin(\theta)}     μkmg=0.3×196 N=58.8 N\mu_k m g = 0.3 \times 196\,\text{N} = 58.8\,\text{N}     cos⁡(20∘)+0.3sin⁡(20∘)=0.9397+0.3(0.3420)=0.9397+0.1026=1.0423\cos(20^\circ) + 0.3 \sin(20^\circ) = 0.9397 + 0.3(0.3420) = 0.9397 + 0.1026 = 1.0423     T=58.8 N1.0423=56.4 NT = \frac{58.8\,\text{N}}{1.0423} = 56.4\,\text{N}

Newton's Third Law and Compound Multi-Body Systems

  • Internal Force Cancellation in Compound Objects:

    • Compound objects composed of multiple interacting parts can be analyzed either as individual separate objects or as a single combined system.

    • Demonstration example: Holding a composite object made of two magnets (e.g., pink and green magnets):

    • Forces on pink magnet: Hand on pink, gravity on pink, green on pink.

    • Forces on green magnet: Gravity on green, pink on green.

    • When evaluating translational equilibrium of the total combined object, the internal interaction forces (Fpink on green\mathbf{F}_{\text{pink on green}} and Fgreen on pink\mathbf{F}_{\text{green on pink}}) cancel out identically due to Newton's Third Law (FA on B=−FB on A\mathbf{F}_{\text{A on B}} = -\mathbf{F}_{\text{B on A}}).

  • Suspended Ball and Floor Block Equilibrium Problem:

    • System parameters:

    • Ball of mass M1=3 kgM_1 = 3\,\text{kg} suspended by a rope at an angle θ=20∘\theta = 20^\circ with the vertical.

    • Block of mass M2=5 kgM_2 = 5\,\text{kg} sitting on a rough horizontal surface with coefficient of static friction μs=0.3\mu_s = 0.3

    • Sides of the block are vertical, restricting any contact force between the ball and block to be strictly horizontal.

  • Equilibrium Analysis of Ball (M1=3 kgM_1 = 3\,\text{kg} ):

    • Gravitational force on ball: Fg1=−(3 kg)(9.8 m/s2)k^=−29.4 Nk^\mathbf{F}_{g1} = -(3\,\text{kg})(9.8\,\text{m/s}^2)\hat{k} = -29.4\,\text{N}\hat{k}.

    • Rope tension T\mathbf{T} acts at an angle of 110∘110^\circ relative to the horizontal x-axis (20∘20^\circ relative to the vertical z-axis).

    • Vertical equilibrium equation:     Tcos⁡(20∘)−M1g=0T \cos(20^\circ) - M_1 g = 0     Tcos⁡(20∘)=29.4 N  ⟹  T=29.4 Ncos⁡(20∘)=31.3 NT \cos(20^\circ) = 29.4\,\text{N} \implies T = \frac{29.4\,\text{N}}{\cos(20^\circ)} = 31.3\,\text{N}

    • Horizontal equilibrium equation:     Tcos⁡(110∘)+Fblock on ball=0T \cos(110^\circ) + F_{\text{block on ball}} = 0     (31.3 N)cos⁡(110∘)+Fblock on ball=0(31.3\,\text{N}) \cos(110^\circ) + F_{\text{block on ball}} = 0     −10.7 N+Fblock on ball=0  ⟹  Fblock on ball=10.7 N-10.7\,\text{N} + F_{\text{block on ball}} = 0 \implies F_{\text{block on ball}} = 10.7\,\text{N}

    • The block exerts a force of 10.7 N10.7\,\text{N} to the left (−10.7 Ni^-10.7\,\text{N}\hat{i}) on the ball.

  • Equilibrium Analysis of Block (M2=5 kgM_2 = 5\,\text{kg} ):

    • Force exerted by ball on block:

    • By Newton's Third Law, Fball on block=−Fblock on ball=10.7 Ni^\mathbf{F}_{\text{ball on block}} = -\mathbf{F}_{\text{block on ball}} = 10.7\,\text{N}\hat{i}.

    • The magnitude of force exerted by the ball on the block is 10.7 N10.7\,\text{N}.

    • Normal force on block:     FN2=M2g=(5 kg)(9.8 m/s2)=49.0 NF_{N2} = M_2 g = (5\,\text{kg})(9.8\,\text{m/s}^2) = 49.0\,\text{N}

    • Maximum static friction capacity:     fs,max=μsFN2=0.3×49.0 N=14.7 Nf_{s,\text{max}} = \mu_s F_{N2} = 0.3 \times 49.0\,\text{N} = 14.7\,\text{N}

    • Actual friction force exerted by floor on block:

    • To maintain horizontal equilibrium: fs=Fball on block=10.7 Nf_s = F_{\text{ball on block}} = 10.7\,\text{N}.

    • Since 10.7 N≤14.7 N10.7\,\text{N} \le 14.7\,\text{N}, the block remains in static equilibrium.

    • The surface exerts a static friction force of 10.7 N10.7\,\text{N} directed to the left (−10.7 Ni^-10.7\,\text{N}\hat{i}).