Pseudo force, gravitation and orbital motion, variation of g with altitude and depth, and turning forces (torque).
Newton's laws work without modification in inertial frames: frames at rest or moving with constant velocity. In an accelerating frame, an observer may introduce a pseudo force to describe apparent motion. It does not arise from a physical interaction.
Fpseudo = −m aframe. The negative sign indicates that the pseudo force acts opposite to the frame's acceleration.
Answer: Newton's laws are valid without modification in an inertial frame of reference.
Answer: A pseudo force is an apparent force introduced when describing motion from an accelerating frame. Its formula is Fpseudo = −m aframe.
Fpseudo = −ma = −60 × 4.5 = −270 N.
Answer: Magnitude = 270 N downward, opposite to the lift's acceleration.
Answer: An inertial frame has zero acceleration, so Fpseudo = −m × 0 = 0.
The Sun's gravitational force provides the centripetal force that continuously changes Earth's velocity direction and keeps it in orbit. Earth's tangential velocity and inertia prevent it from falling straight into the Sun.
Astronauts and their spacecraft are both in continuous free fall around Earth. They appear weightless because they accelerate together; gravity has not disappeared.
In a vacuum there is no air resistance. Since F = mg and a = F/m, a = mg/m = g. The mass cancels, so objects have the same gravitational acceleration at the same location.
gh = g [R/(R + h)]². As altitude increases, the distance from Earth's centre increases and g decreases.
For the uniform-density Earth model, gd = g(1 − d/R). In this simplified model, g decreases towards the centre and becomes zero at the centre.
Answer: In the chapter's simplified model, g is maximum at Earth's surface and decreases above and below it.
Answer: It becomes zero in the idealised spherical model because the gravitational pulls balance.
Using g = 9.8 m/s² and gh = g[R/(R+h)]²:
gh = 9.8 × [6400/6800]² ≈ 8.68 m/s².
g/2 = g(1 − d/R), so 1/2 = 1 − d/R and d = R/2.
Answer: Half the Earth's radius; for R = 6400 km, 3200 km.
Answer: Above Earth, distance from its centre increases. Below the surface, the enclosed mass contributing to the net gravitational pull decreases towards the centre in the uniform-density model.
Torque is the turning effect of a force about a pivot.
τ = F × d × sin θ
F is force, d is the distance from pivot to the point of application, and θ is the angle between the force and lever arm. SI unit: N m.
Answer: The handle is farther from the hinge, giving a larger lever arm and greater torque for the same perpendicular force.
Answer: Torque is maximum at 90°. If force is parallel or antiparallel to the wrench, sin θ = 0 and torque is zero.
τ = Fd. The second torque is F × 0.8, while the first is F × 0.2.
Answer: The student pushing at 80 cm produces four times the torque.
Answer: Yes. If the line of action passes through the pivot, the perpendicular distance is zero. For example, pushing a door exactly at its hinge produces no turning effect.
Each torque has magnitude 10 × 0.5 = 5 N m, but the directions are opposite. Net torque = 5 − 5 = 0 N m.
Answer: The torques balance, so there is no net turning effect.
Answer: A longer spanner increases the lever arm, so it produces greater torque with the same force.
τ = Fd sin θ = 20 × 0.8 × sin θ = 16 sin θ.
Note: The altitude and depth equations are simplified textbook models. Earth's actual density is not uniform.