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Space, Time and Motion - Forces and Newton's Laws

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Newton's First Law (Law of Inertia): An object continues in a state of rest or uniform motion in a straight line unless acted upon by a non-zero net external force (Fnet=0F_{net} = 0).

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Newton's Second Law: The net force acting on an object is equal to the rate of change of its linear momentum: F=ΔpΔtF = \frac{\Delta p}{\Delta t}. For constant mass, this simplifies to F=maF = ma.

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Newton's Third Law: When two bodies interact, the force exerted by body AA on body BB is equal in magnitude and opposite in direction to the force exerted by body BB on body AA. These forces act on different objects.

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Translational Equilibrium: An object is in translational equilibrium if the vector sum of all forces acting on it is zero (∑F⃗=0\sum \vec{F} = 0). It is either at rest or moving with constant velocity.

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Weight (WW): The force of gravity acting on an object, calculated as W=mgW = mg, where g≈9.81 m s−2g \approx 9.81 \text{ m s}^{-2} on Earth.

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Friction: The force that opposes relative motion. Static friction (Ff≤μsRF_f \leq \mu_s R) acts when there is no relative motion; Kinetic friction (Ff=μkRF_f = \mu_k R) acts when there is sliding motion.

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Free-Body Diagrams (FBD): A diagram representing all the external forces acting on a single body, shown as vectors originating from a point or the center of mass.

📐Formulae

Fnet=maF_{net} = ma

p=mvp = mv

F=ΔpΔtF = \frac{\Delta p}{\Delta t}

W=mgW = mg

Ff≤μsRF_f \leq \mu_s R

Ff=μkRF_f = \mu_k R

💡Examples

Problem 1:

A wooden block of mass m=4.0 kgm = 4.0 \text{ kg} is pulled across a horizontal surface by a constant horizontal force of 20 N20 \text{ N}. The coefficient of kinetic friction μk\mu_k between the block and the surface is 0.300.30. Calculate the acceleration of the block. (Take g=9.81 m s−2g = 9.81 \text{ m s}^{-2})

Solution:

  1. Identify the forces: Pulling force Fp=20 NF_p = 20 \text{ N}, Normal reaction R=mg=4.0×9.81=39.24 NR = mg = 4.0 \times 9.81 = 39.24 \text{ N}, Friction force Ff=μkR=0.30×39.24=11.772 NF_f = \mu_k R = 0.30 \times 39.24 = 11.772 \text{ N}.
  2. Use Newton's Second Law: Fnet=Fp−Ff=maF_{net} = F_p - F_f = ma.
  3. Substitute values: 20−11.772=4.0×a20 - 11.772 = 4.0 \times a.
  4. 8.228=4.0a  ⟹  a=8.2284.0=2.057 m s−28.228 = 4.0a \implies a = \frac{8.228}{4.0} = 2.057 \text{ m s}^{-2}.

Explanation:

The net force is the difference between the applied pulling force and the opposing kinetic friction. The acceleration is then found by dividing this net force by the mass of the block.

Problem 2:

An object of mass 10 kg10 \text{ kg} is suspended by two strings at angles of 30∘30^\circ and 60∘60^\circ to the horizontal. Calculate the tension in each string when the system is in equilibrium.

Solution:

  1. Weight W=10×9.81=98.1 NW = 10 \times 9.81 = 98.1 \text{ N} downwards.
  2. Horizontal equilibrium: T1cos⁡(30∘)=T2cos⁡(60∘)  ⟹  T132=T212  ⟹  T2=T13T_1 \cos(30^\circ) = T_2 \cos(60^\circ) \implies T_1 \frac{\sqrt{3}}{2} = T_2 \frac{1}{2} \implies T_2 = T_1 \sqrt{3}.
  3. Vertical equilibrium: T1sin⁡(30∘)+T2sin⁡(60∘)=WT_1 \sin(30^\circ) + T_2 \sin(60^\circ) = W.
  4. Substitute T2T_2: T1(0.5)+(T13)(32)=98.1T_1 (0.5) + (T_1 \sqrt{3})(\frac{\sqrt{3}}{2}) = 98.1.
  5. 0.5T1+1.5T1=98.1  ⟹  2T1=98.1  ⟹  T1=49.05 N0.5 T_1 + 1.5 T_1 = 98.1 \implies 2T_1 = 98.1 \implies T_1 = 49.05 \text{ N}.
  6. T2=49.05×3≈84.96 NT_2 = 49.05 \times \sqrt{3} \approx 84.96 \text{ N}.

Explanation:

Since the object is in equilibrium, the sum of horizontal components must be zero, and the sum of vertical components must equal the weight of the object.

Forces and Newton's Laws - Revision Notes & Key Formulas | IB Grade 11 Physics