Chapter 7 Vectors Kinematics Gravity and Force Balance

Year 1 · Physics · G113-L1

Apply vector components, Newton’s laws, gravity and equilibrium to a worked slope-force problem and geological measurements with clear units.

Wavelength, amplitude and travel-time relationships in a simple wave model. Original Geology Hub teaching diagram.

Learning objectives: Resolve vectors, distinguish displacement from distance, apply Newton’s laws, and check equilibrium and gravitational calculations using consistent units.

Measurements need magnitude, direction and a reference

A scalar has magnitude; a vector also has direction. Mass and temperature are scalars, whereas displacement, velocity and force are vectors. Walking three metres east and three metres west gives six metres of travel but zero net displacement. The distinction matters when geological displacement is reconstructed from markers: the length of a path and the final separation need not agree.

Choose coordinates before calculating. If a displacement of length L makes an angle θ above the horizontal, its horizontal and vertical components are L cos θ and L sin θ. Components can be added algebraically once their signs are defined. The dot product measures alignment between vectors and underlies work; the cross product describes a perpendicular vector and underlies torque. A scalar field assigns a number to each location, while a vector field assigns a magnitude and direction. Gradient describes spatial change of a scalar, and divergence describes local spreading or convergence of a vector field.

Motion connects position, velocity and acceleration

Average velocity is displacement divided by elapsed time. Instantaneous velocity is the rate at which position changes at one moment. Acceleration is the rate of change of velocity, which can occur through changing speed, changing direction or both. Under constant acceleration, v = u + at and displacement s = ut + ½at². These equations require a defined positive direction and constant acceleration over the interval.

Gravity provides a familiar approximation near Earth’s surface. A freely falling body accelerates downward if air resistance is negligible. Its mass is an amount of matter; weight is gravitational force. Local gravitational acceleration varies with location, altitude and Earth’s rotation. In introductory problems, using g = 9.8 metres per second squared is often adequate, but a precision geophysical survey must account for those variations.

Forces and equilibrium

Newton’s first law describes constant velocity when net force is zero. The second relates net force to mass and acceleration. The third states that interacting bodies exert equal and opposite forces on each other. Those paired forces act on different bodies, so they do not cancel on a free-body diagram of just one body. Draw only forces acting on the object under study.

For a rigid body in static equilibrium, net force and net torque must both vanish. A force applied farther from a pivot can produce more torque, depending on its perpendicular lever arm. The centre of gravity is a useful location for representing the resultant weight. Projectile motion separates horizontal and vertical components when drag is ignored. In circular motion, acceleration points toward the centre; “centripetal” names this role rather than a new kind of physical interaction.

Fields describe quantities throughout space

A temperature field assigns one scalar value to each point. Its gradient points toward the greatest local increase and has units of temperature per distance. A groundwater-flow field instead assigns a vector at each point. Divergence measures local net outward spreading in an appropriate mathematical model; it is not simply the speed of the flow. The distinction matters because neighbouring locations can have equal speeds but different directions, producing a different spatial pattern.

Products answer different geometric questions

The dot product of two vectors is a scalar involving their magnitudes and the cosine of their included angle. Work uses this relationship because only the force component along displacement contributes. The cross product has a direction perpendicular to both original vectors and a magnitude involving their sine. Torque uses the cross product of position relative to a pivot and force. A force directed through the pivot has no perpendicular lever arm and therefore produces no torque about that pivot.

Separate projectile motion into components

Under negligible air resistance, horizontal velocity stays constant while gravity changes vertical velocity. A fictional particle launched horizontally at 4 metres per second travels 8 horizontal metres in two seconds. Over that interval its vertical fall is approximately ½ × 9.8 × 2² = 19.6 metres if it remains freely falling. These independent component calculations describe one curved path. They do not imply that a rockfall near real terrain has no collisions or drag.

Gravity connects falling bodies and satellites

Newtonian attraction between masses decreases with the square of their separation in the point-mass model. A satellite can orbit because gravity supplies inward acceleration while its tangential motion continually carries it forward. It is not outside gravity merely because it is in orbit. Increasing altitude generally reduces gravitational acceleration, while Earth’s rotation affects measured effective gravity. Mass describes inertia as well as gravitational interaction, but weight is the force associated with the local gravitational situation.

Equilibrium depends on force placement

Place a hypothetical 10-newton weight 0.6 metres from a support. Its torque magnitude is 6 newton metres when the force is perpendicular to the lever arm. A 20-newton force at 0.3 metres on the opposite side can balance that torque. A complete static solution must also balance vertical forces, including the support reaction. This explains why locating a centre of gravity matters for a rigid body: equal total weights placed differently can have different overturning tendencies. The example teaches mechanics rather than a stability assessment of an actual rock mass.

Worked slope calculation

Consider a hypothetical 20-kilogram block on a smooth plane inclined at 30 degrees. Weight is 20 × 9.8 = 196 newtons. Resolve it parallel and perpendicular to the plane: 196 sin 30° = 98 newtons downslope and 196 cos 30° ≈ 170 newtons into the plane. With no friction or other downslope force, acceleration is 98/20 = 4.9 metres per second squared.

The same calculation is not a real slope-stability assessment because natural slopes involve joints, fluid pressures, heterogeneous material and complex geometry. Its learning result is a correctly labelled free-body diagram and a checked force balance. Change the angle to zero: the downslope component should vanish. Such limiting checks often reveal an incorrect sine, cosine or sign before it reaches a report.

Self-check

Can speed remain constant while acceleration is nonzero?

Yes. Circular motion changes velocity direction even when speed stays constant.

Do Newton’s third-law forces cancel on one block?

No. The paired forces act on different objects.

What happens to the downslope weight component on a horizontal plane?

It becomes zero because sin 0° is zero.

Sources and further study

Topic framework: the supplied University of Peshawar BS Geology curriculum, session 2004–05 onward. This independently authored learning resource is not affiliated with the university and does not confer a degree or accreditation.

Chapter 7 multiple choice questions

Choose one best answer for each question. Open each answer panel after making your choice.

A student walks 3 m east and then 3 m west. What are the total distance and net displacement?
1. 6 m distance and 0 m displacement
2. 0 m distance and 6 m east displacement
3. 6 m distance and 6 m west displacement
4. 3 m distance and 3 m east displacement
Answer and explanation
A. 6 m distance and 0 m displacement Distance totals path length, while displacement depends only on the change from starting to ending position.
A 20 kg block rests on a frictionless 30° slope. With g = 9.8 m/s², what is the downslope component of its weight?
1. 196 N
2. 170 N
3. 98 N
4. 20 N
Answer and explanation
C. 98 N The parallel component is mg sin 30° = 20 × 9.8 × 0.5 = 98 N.
For the same frictionless block, what downslope acceleration follows?
1. 9.8 m/s²
2. 4.9 m/s²
3. 0.5 m/s²
4. 98 m/s²
Answer and explanation
B. 4.9 m/s² Net downslope force divided by mass is 98/20 = 4.9 m/s².
A vehicle moves at constant speed around a circular path. Why does it still accelerate?
1. Its mass necessarily increases.
2. Its speed must be changing despite the stated condition.
3. Every constant-speed motion has zero acceleration.
4. Its velocity direction changes.
Answer and explanation
D. Its velocity direction changes. Velocity is a vector, so changing direction produces acceleration even if speed stays constant.
A rigid body remains at rest without rotating. Which conditions describe static equilibrium?
1. Net force is zero, even when a nonzero torque remains.
2. Net torque is zero, even when a nonzero force remains.
3. Both net force and net torque are zero.
4. Clockwise torques vanish, while anticlockwise torques can remain.
Answer and explanation
C. Both net force and net torque are zero. Static equilibrium requires no resultant translational force and no resultant rotational influence.

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