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A-Level Types of force, turning effects and conditions for equilibrium
What the A-Level syllabus expects for Types of force, turning effects and conditions for equilibrium, and how to practise it.
What the syllabus expects
- Account for the force felt by a mass, by a charge and by a current-carrying wire when each sits in a gravitational, electric or magnetic field, as the situation demands
- Describe, without calculation, everyday forces such as the normal contact force, buoyancy (upthrust), friction and drag like air resistance
Scope: the coefficients of friction and of viscosity are not examined - Bring Hooke's law to bear on fresh scenarios and on related calculations
Scope: F = kx, with k the force constant - State, then put to use, both the turning moment produced by a single force and the torque produced by a couple
- Recognise a couple as two forces whose only tendency is to spin an object
- Treat a body's weight as though it all pressed down through one spot, its centre of gravity
- Put the principle of moments to work on unfamiliar setups and on problem solving
- Grasp that a system rests in equilibrium precisely when both its net force and its net torque vanish
- Draw free-body diagrams and closed vector triangles to depict the forces on objects balanced against both turning and sliding
How it's examined
About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (2 marks)
A uniform rod of mass m leans so that it just touches a hoop tangentially at point A, making an angle θ with the horizontal. Demonstrate, setting out your reasoning clearly, that the normal contact force N_A at A is equal to (1/2) m g cos θ, where g is the acceleration of free fall.
Show the worked answer
Model the rod as a uniform rigid body of weight mg, resting with its lower end at B and inclined at angle θ to the horizontal, its upper end just touching the hoop tangentially at A. Step 1 - direction of the contact force. Because the rod touches the hoop TANGENTIALLY, the common tangent at the point of contact is the rod itself. The contact (normal) force always acts perpendicular to the common tangent, so N_A must act perpendicular to the rod. Step 2 - the other forces. The weight mg acts vertically downwards at the midpoint G of the rod (uniform rod). At the foot B the ground supplies a reaction (a normal component and friction). We do not need to know these if we take moments about B, because their lines of action pass through B. Step 3 - take moments about B. Two forces have a moment about B: N_A and the weight. • N_A is perpendicular to the rod, so its moment arm about B is simply the distance BA along the rod. Moment = N_A · (BA). • The weight is vertical, so its moment arm about B is the HORIZONTAL distance from B to G. Since G is the midpoint, BG = ½·BA, and the horizontal projection is (½·BA)·cos θ. Moment = mg · (½ BA cos θ). Step 4 - balance. For equilibrium the moments balance: N_A · (BA) = mg · (½ BA cos θ). The length BA cancels, giving N_A = ½ mg cos θ. □ The result is independent of the rod's length precisely because the weight acts at the midpoint (half-way along BA), so the geometric factor cancels.
More A-Level H2 Physics topics
Physical quantities, units, measurement uncertainty and vector basics · Kinematics, uniformly accelerated motion, momentum and Newton's laws · Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency · Falling freely, the gravitational potential energy of a uniform field, and how air resistance changes the motion · Impulse and the conservation of momentum and energy · Kinematics of uniform circular motion and centripetal acceleration · all of A-Level H2 Physics