Mechanics | Statics and Dynamics from scratch

Explore the composition of forces and the law of parallelogram to determine the resultant of two or more forces, with magnitude r = sqrt(p^2+q^2+2pq cos alpha) and direction theta.

Apply the resultant formula r = sqrt(p^2 + q^2 + 2 p q cos alpha) to calculate r for the three examples given, including 5 N and sqrt(364) N.

Find the resultant of two forces, 10 N and 15 N at 60 degrees, using r = sqrt(p^2 + q^2 + 2pq cos alpha), giving r = 5 sqrt(19) N.

Compute the resultant of two forces, 3 N and 1 N, at 60 degrees. Determine magnitude sqrt(13) N and direction with theta = arctan(Q sin alpha/(P+Q cos alpha)).

Apply the perpendicular-force resultant rule to p+q and p−q, showing their resultant magnitude equals sqrt(p^2+q^2) when alpha is 90 degrees and cos alpha equals zero.

Demonstrate that the resultant of two equal forces P at 120 degrees equals P by applying the resultant formula and substituting cos 120 degrees = -1/2.

Compute the angle between two equal forces using the resultant, with R^2 = (2 - sqrt(3)) P^2, yielding cos alpha = - sqrt(3)/2 and alpha = 150 degrees.

Show how two equal forces inclined at 2 theta yield a resultant, twice the magnitude of the resultant at 2 phi. Derive cos theta = 2 cos phi.

Find the angle between two equal forces p when their resultant is p, using R^2 = P^2 + Q^2 + 2PQ cos alpha; conclude alpha is 120 degrees.

Show that equating the two expressions for the resultant leads to cos alpha = 1, so alpha is zero and the resultant equals p plus q.

Determine the angle alpha between forces p+q and p−q so their resultant has magnitude sqrt(p^2+q^2) using squaring and cos alpha to solve for alpha.

Apply the law of cosines to two forces of 50 and 120 N balanced by 130 N, and determine that their angle is 90 degrees, illustrating vector addition in statics.

Determine alpha, the angle between p plus q and p minus q, so that the resultant equals sqrt(p^2+3q^2), giving cos alpha equals minus one half.

Two equal forces on a particle produce a resultant with r^2 equals p^2 plus p^2 plus 2 p^2 cos alpha; solving shows alpha, angle between the forces, equals 60 degrees.

Apply the law of cosines to two forces with magnitudes 0.03 and 0.02, using the doubling condition to derive that the angle between them is 120 degrees.

Derive the resultant of two forces p and q as r, compute the resultant s when p doubles (q fixed), showing s^2 = 0.02^2 + 2 r^2 - q^2.

Compute the angle between two ten-newton forces balanced by a five-newton resultant using the law of cosines, giving cos inverse of -7/8 for alpha.

Two forces at a point yield a maximum resultant of four newtons; at right angles it's three newtons, giving p and q as 2±1/√2.

The lecture derives that the resultant of two forces with magnitudes r and s at angle alpha equals sqrt(r^2 cos^2(alpha/2) + s^2 sin^2(alpha/2)).

Find the greatest and least resultant of forces p and q at 60 degrees with p = 10 N and q = 20 N, giving r = 10 sqrt7 N.

Resolve a single force into two components using parallelogram construction, equating the resultant to the diagonal. Apply the sine rule to obtain P and Q with angles alpha and beta.

Decompose a 10 N force into components p and q at 30 and 45 degrees using the sine rule. Find p = 20/(√3+1) N and q = 10√2/(√3+1) N.

Compute the components of a force of 32 newtons in directions at 30 and 60 degrees. Derive p and q as 16 square root three newtons and 16 newtons, respectively.

Explain how to resolve a force into perpendicular components along horizontal and vertical axes, deriving horizontal component f cos theta and vertical component f sine theta.

Resolve a 36 N force given a component of 18 sqrt(3) N; cos theta equals sqrt(3)/2, theta is 30 degrees, and the other resolved component is 18 N.

Determine the angle and components of a 32 newton force whose projection is half its magnitude. The angle is 60 degrees, giving the other component 16 root three newtons.

Compute the horizontal and vertical components of a 10-newton force at 30 degrees, yielding 10 cos 30 degrees (five root three newtons) and 10 sin 30 degrees (five newtons).

Resolve a 16 newton force into components by setting 16 cos θ = 8√3 to yield θ = 30 degrees, and 16 sin θ = 8 N as other component.

Resolve a 42 newton force into components: use cos theta = (21 sqrt 2)/42 to get theta = 45, and 42 sin 45 = 21 sqrt 2 newton.

Explore Lami's theorem for equilibrium of concurrent forces: three forces at a point are proportional to the sine of the angle between the other two, and its converse.

Apply Lami's theorem to three forces in equilibrium and determine their magnitudes p:q:r = 1:1:√3 from angles 60° and 150°.

Apply lami's theorem to three coplanar forces in equilibrium. Compute the angles, establish a 90 degree angle between p and r, and obtain p:q:r = 1:2:√3.

Apply Lami's theorem to three forces in equilibrium with 120-degree angles between each pair. Show that P equals Q equals R.

Apply Lami's theorem to three equal forces in equilibrium, showing that the angles between them are equal and sum to 360 degrees, yielding each angle as 120 degrees.

From mechanics, a 40 N weight suspended by two strings at 30° and 60° from the vertical uses Lami's theorem to find tensions T1 and T2 in equilibrium.

Define key dynamics concepts, including velocity, acceleration, speed, and displacement, and distinguish between vector and scalar quantities, distance versus displacement, and instantaneous versus average speed.

Calculate the bicycle’s average speed for a 50 km journey split into four phases; add phase times to five hours and divide 50 km by five hours.

Compute the distance: a particle moves 3 m/s for 3 s, then 4 m/s for 2 s at a right angle, giving a distance of 17 m from the start.

An eastward leg from A to B at 4 km/h, then north to C at 5 km/h, ends at A. With zero displacement, average velocity equals zero.

Derive velocity and acceleration from the distance-time relation x = 20 t^2 + 50 t + 19; at t = 3 s, v = 170 cm/s and a = 40 cm/s^2.

example-5 shows a particle with x=63t-6t^2-t^3; velocity v=63-12t-3t^2, giving v(2)=27 m/s, stopping at t=3 s with x(3)=108 m.

derive acceleration from the velocity-distance relation v = 10 + s/15 by differentiating to dv/ds = 1/15, compute a = v dv/ds at s = 900 cm, giving 14/3 cm/s^2.

Compute x(t)=2t^3-9t^2+5t+8 to obtain v=dx/dt and a=d^2x/dt^2; set a=0 to find t=3/2 s, then v=-8.5 m/s.

Apply v^2 = u^2 + 2 f s to a uniformly accelerated automobile between two poles, with u = 10 km/h and halfway velocity v = 5√10 km/h.

Solve a uniformly accelerated motion problem from the lecture: given u=25 m/s, v=55 m/s in 10 s, find a and distance s using v^2 = u^2 + 2 a s.