Cards In This Deck

• $\text{What is the formula for Weight (}W\text{) derived from SI base units?}$
  $$\begin{gathered} W=mg \\ \text{Where }m\text{ is mass (}kg\text{) and }g\text{ is acceleration due to gravity (}m{s}^{-2}\text{).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{The normal reaction }R\text{ on an object of mass }m\text{ resting on an inclined plane is always equal to }mg\cos \alpha \text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }R=mg\cos \alpha \text{ only if there are no other external forces acting perpendicular to the plane.} \\ \text{If an external force }P\text{ pushes into the plane, }R=mg\cos \alpha +P\sin \theta \text{ (depending on angles).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{The friction force }F\text{ acting on a rough surface is always calculated using }F= \\ \mu R\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }F= \\ \mu R\text{ only applies when the object is moving or in limiting equilibrium (on the point of moving).} \\ \text{If the object is stationary and not on the point of slipping, }F< \\ \mu R\text{ (Friction equals the applied force).} \end{gathered}$$
• $\text{A particle is in equilibrium on a rough inclined plane. You need to determine the direction of the friction force. How do you decide?}$
  $$\begin{gathered} \text{Friction acts in the direction opposite to the potential motion.} \\ \text{Ask yourself: "If the surface were smooth, which way would the particle move?" Friction points the other way.} \end{gathered}$$
• $\text{A particle is in equilibrium under 3 forces: }A\text{, }B\text{, and }C\text{, with angles }\alpha \text{, }\beta \text{, and }\gamma \text{ opposite to them respectively. How do you apply Lami’s Theorem?}$
  $\frac{A}{\sin \alpha }=\frac{B}{\sin \beta }=\frac{C}{\sin \gamma }$
• $\text{You are solving a statics problem with exactly three non-parallel coplanar forces acting on a particle. Which method is often faster than resolving forces?}$
  $$\begin{gathered} \text{Triangle of Forces (or Lami’s Theorem).} \\ \text{Sketch a closed vector triangle. You can then use the Sine Rule or Cosine Rule to find unknown magnitudes or angles without resolving into }x\text{ and }y\text{ components.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{When drawing a Triangle of Forces for a particle in equilibrium, the arrows representing the forces must all point away from the same vertex.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Method: The arrows must follow a cyclic order (nose-to-tail) around the triangle. } \\ \text{If you start at a point and follow the arrows, you should arrive back at the start (representing a resultant force of zero).} \end{gathered}$$
• $\text{What are the two necessary conditions for a rigid body to be in static equilibrium?}$
  $$\begin{gathered} \text{1. The resultant force in any direction is 0 N.} \\ \text{2. The resultant moment about any point is 0 Nm.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{The Normal Reaction }R\text{ on an object of mass }m\text{ sitting on an inclined plane (angle }\alpha \text{) is always }mg\cos \alpha \text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: }R=mg\cos \alpha \text{ is only true if no other external forces have a component perpendicular to the plane.} \\ \text{If there is an external force }P\text{ pulling the object at an angle away from the plane, }R\text{ will be less than }mg\cos \alpha \text{. Always resolve perpendicular to the plane to find }R\text{.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{The Normal Reaction force (}R\text{) always acts vertically upwards.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: It acts perpendicular (at }90^{\circ }\text{) to the surface.} \end{gathered}$$
• $\text{What is the standard first step when resolving forces for a particle on an inclined plane?}$
  $$\begin{gathered} \text{Resolve forces in two perpendicular directions:} \\ \text{1. Parallel to the line of greatest slope (}\nearrow \text{ or }\swarrow \text{).} \\ \text{2. Perpendicular to the plane (}\nwarrow \text{ or }\searrow \text{).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If a question states an object has a mass of }10kg\text{, its weight is }10kg\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: Weight is a force measured in Newtons (}N\text{). } \\ W=mg=10\times 9.8=98N \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If }{F}_{max}=20N\text{ and the applied force trying to move the object is only }10N\text{, then the friction force is }20N\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: The friction force is }10N\text{.} \\ \text{In non-limiting equilibrium (}F< \\ \mu R\text{), friction is ’lazy’—it only exerts exactly enough force to oppose the motion and maintain equilibrium, up to the maximum limit.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{For a particle in equilibrium on a rough inclined plane, the friction force is always equal to }F= \\ \mu R\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Understanding: }F= \\ \mu R\text{ only applies when the particle is in limiting equilibrium (on the point of moving) or moving.} \\ \text{If the particle is stationary but not on the point of slipping, then }F< \\ \mu R\text{ (friction only matches the component of the applied force/gravity to maintain equilibrium).} \end{gathered}$$
• $\text{What condition must be met for a particle to remain at rest on a rough inclined plane with angle }\alpha \text{ (under gravity only)?}$
  $$\begin{gathered} \tan \alpha \le \\ \mu \\ \text{(The component of weight down the slope }mg\sin \alpha \text{ must be less than or equal to the maximum friction } \\ \mu mg\cos \alpha \text{)} \end{gathered}$$
• $\text{Under what exact condition is a particle said to be in Limiting Equilibrium?}$
  $$\begin{gathered} \text{1. The particle is in equilibrium (stationary).} \\ \text{2. The friction force has reached its maximum value: }F= \\ \mu R\text{.} \\ \text{3. The particle is on the point of moving (slipping).} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If a particle is in ’limiting equilibrium’, the net force acting on it is } \\ \mu R\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: The net force is zero (}0\text{).} \\ \text{’Equilibrium’ means acceleration is zero. ’Limiting’ implies the friction magnitude is currently at its maximum, }{F}_{fric}= \\ \mu R\text{, but forces still balance out.} \end{gathered}$$
• $$\begin{gathered} \text{True or False? } \\ \text{If the resultant moment on a body is zero, the body must be in equilibrium.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct: Zero resultant moment only prevents rotation. For full equilibrium, the resultant force must ALSO be zero to prevent translation.} \end{gathered}$$
• $\text{In a problem involving a ’light smooth pulley’, what two assumptions can you make?}$
  $$\begin{gathered} \text{1. Light: The pulley has no mass (ignore gravity acting on the pulley itself unless asked for force on axle).} \\ \text{2. Smooth: The tension in the string is the same magnitude on both sides of the pulley.} \end{gathered}$$
• $\text{What is the mathematical consequence of modelling a string (or pulley) as light?}$
  $\text{The tension is the same at both ends of the string (or on either side of the pulley).}$
• $$\begin{gathered} \text{A smooth bead is threaded onto a string. } \\ \text{What can we say about the tension in the string on either side of the bead?} \end{gathered}$$
  $\text{The tension is the same on both sides.}$
• $$\begin{gathered} \text{True or False?} \\ \text{If a light inextensible string passes through a smooth bead, the tension in the string is the same on both sides of the bead.} \end{gathered}$$
  $$\begin{gathered} \text{✅ True} \\ \text{Because the bead is smooth, it cannot exert a tangential force on the string to change the tension. Therefore, magnitude }T\text{ is constant throughout the single string.} \end{gathered}$$