Force solutions

Take right as +x and up as +y. The upper angled force is (−45 cos 30°, +45 sin 30°) N; the lower angled force is (+45 cos 45°, +45 sin 45°) N.

Why: Fixing right and up as positive makes the signs consistent. Each angled force's horizontal projection uses cosine, while its vertical projection uses sine.

Add x components: ΣFₓ = 55 − 50 − 45 cos 30° + 45 cos 45° = −2.15134 N.

Why: Add every horizontal force on the box. The upper leftward force contributes negatively, and the lower rightward force contributes positively.

Add y components: ΣFᵧ = −15 − 98 + 98 + 45 sin 30° + 45 sin 45° = +39.3198 N.

Why: Add every vertical force on the box. The 98 N pair cancels, but the extra 15 N downward force still belongs in the sum.

Divide each component by 10 kg: a = (−0.215134, +3.93198) m/s².

Why: Newton's second law applies to each axis: ax = ΣFx/m and ay = ΣFy/m. The same 10 kg mass goes in both denominators.

Magnitude = √(aₓ² + aᵧ²); atan2(aᵧ,aₓ) gives the direction in quadrant II.

Why: The acceleration components are perpendicular, so Pythagoras gives their combined size. Negative ax and positive ay put the direction in quadrant II.

At equilibrium the upward spring force balances weight: kx = mg.

Why: The hanging object is at rest, so its net force is zero. The ideal spring's upward kx must equal the downward weight.

The extension is x = (20 kg)(9.8 m/s²)/(96 N/m) = 2.04167 m.

Why: Solve the force balance for x = mg/k. The result is the spring's added length, and N divided by N/m leaves metres.

The requested anchor-to-object distance includes the 0.50 m unstretched length: L = 0.50 + 2.04167.

Why: The anchor-to-object distance is the whole stretched spring, not just its extension. Add the original 0.50 m length back in.

Convert 5.8 Mg to 5800 kg and 11.6 kN to 11 600 N.

Why: Use consistent SI units before applying F = ma: mega means a million grams, and kilo means a thousand newtons.

The required net force is ma = 5800(6) = 34 800 N.

Why: With resistance neglected, the sails supply the entire net force needed to accelerate the 5800 kg ship.

Divide by the force from one sail: n = 34 800/11 600 = 3.

Why: Each sail contributes the same 11 600 N, so divide the total force requirement by one sail's contribution to count the sails.

Choose downward as positive. The net force is mg − D.

Why: Downward-positive makes weight positive and the upward resistance negative. Their difference sets the falling skydiver's acceleration.

At terminal speed, mg − D = 0: 70(9.8) = 8v² + 10.

Why: Terminal speed means the velocity stops changing. Set the net force to zero even though the skydiver is still moving.

Therefore v² = (686 − 10)/8 = 84.5 m²/s².

Why: Subtract the speed-independent 10 N resistance before dividing by the coefficient of v². This isolates the squared speed.

Take the nonnegative square root because the question asks for speed.

Why: Speed is a magnitude, so use the positive root. The physical velocity points downward, as stated in the chosen falling motion.

Vertical acceleration is zero while the box stays on the floor: N + 40 sin 40° − mg = 0. Thus N = 490 − 25.7115 = 464.2885 N.

Why: The box has no vertical acceleration. The pull's upward component helps support it, so the floor's normal force is smaller than mg.

Horizontal balance gives 40 cos 40° − fₖ = ma = 25 N. Thus fₖ = 30.6418 − 25 = 5.6418 N.

Why: Sliding to the right puts kinetic friction to the left. The pull's horizontal component must provide both that friction force and the 25 N net force.

Use fₖ = μₖN: μₖ = 5.6418/464.2885.

Why: The kinetic-friction model gives fₖ = μₖN. Divide by the actual reduced normal force, not by the full weight; the units cancel.

Take uphill for the 2 kg block and downward for the 5 kg block as positive. The rope gives equal acceleration magnitudes. N = 2g cos 30° = 16.9741 N.

Why: The fixed pulley and taut rope tie the two motions together. Choose positive directions that correspond to the same rope motion; the ramp-normal force balances perpendicular weight.

If the ramp block moves uphill, friction points downhill: T − 2g sin 30° − μₖN = 2a, and 5g − T = 5a.

Why: For uphill sliding, friction points downhill along with the ramp component of weight. The hanging block's weight pulls in its positive direction.

Add the equations: a = [5g − 2g sin 30° − μₖ(2g cos 30°)]/7.

Why: Adding the two block equations cancels the internal tension. The remaining external driving force accelerates their combined 7 kg mass.

For μₖ = 0.20, a = (49 − 9.8 − 3.39482)/7 = +5.11503 m/s²; for μₖ = 0.10, a = (49 − 9.8 − 1.69741)/7 = +5.35751 m/s².

Why: Insert each section's coefficient separately because the normal force stays the same while the friction magnitude changes.

If the ramp block instead moves downhill, friction points uphill. Change the friction minus sign to plus in the summed equation; the accelerations are +6.08497 and +5.84249 m/s².

Why: Friction opposes velocity, not acceleration. For downhill sliding it points uphill, adding to the force that gives an uphill acceleration and slows the block.

Assume massless frictionless pulleys and an inextensible rope, so the rope tension is uniform.

Why: An ideal massless rope over frictionless massless pulleys has the same tension throughout; no pulley needs a tension difference to accelerate its rotation.

The fixed pulley geometry gives equal and opposite end accelerations. Choose 10 kg downward and 5 kg upward as positive.

Why: All axles are fixed, so the extra pulleys only redirect the rope. If one end moves down, the other must move up by the same amount.

For the masses: 10g − T = 10a and T − 5g = 5a.

Why: Write F = ma for each mass using its chosen positive direction. Tension opposes the heavier mass's descent but supports the lighter mass's ascent.

Add: (10 − 5)g = (10 + 5)a. Thus a = 49/15 = 3.26667 m/s².

Why: Summing the equations removes the internal rope tension. The difference in weights drives the combined mass.

Let the original shared acceleration be a₀. On A: 40 − 30 = m_Aa₀, so m_Aa₀ = 10 N.

Why: For A, its pull and backward coupling force leave only 10 N of net force. That net force equals mA times the shared original acceleration.

On B: 30 − 20 = m_Ba₀, so m_Ba₀ = 10 N. On C: m_Ca₀ = 20 N. Therefore m_A = m_B and m_C = 2m_A.

Why: B's two tensions leave 10 N, while C receives 20 N. Since all three share a₀, these net-force ratios reveal their mass ratios.

After C separates, a₁ = 40/(m_A+m_B).

Why: Once C leaves, the same external 40 N accelerates only A and B. Their smaller combined mass determines the new acceleration.

The coupling accelerates B alone: T_new = m_Ba₁ = 40 m_B/(m_A+m_B) = 20 N.

Why: The A–B coupling is the only horizontal force on B. With equal remaining masses, it supplies half the 40 N system force.

For the hanging block at rest, T = 5g + F = 49 + F N.

Why: The hanging block has zero acceleration. Its upward tension must balance both its weight and the added downward force.

For the ramp block, downhill weight component = 12g sin 30° = 58.8 N and normal force N = 12g cos 30° = 101.8446 N.

Why: Resolve the ramp block's weight into downhill and ramp-normal components. Only the perpendicular component sets N here.

Maximum static friction magnitude is μₛN = 1.01845 N. Ramp equilibrium is possible if |58.8 − T| ≤ 1.01845.

Why: Static friction adjusts either way along the ramp, but its magnitude cannot exceed μsN. It can balance the weight–tension difference only within that limit.

Substitute T = 49 + F: |9.8 − F| ≤ 1.01845, which gives 8.78155 ≤ F ≤ 10.81845 N.

Why: Replace T with the hanging block's equilibrium expression. Solving both sides of the absolute-value inequality gives the allowed downward-force interval.

At the minimum, friction acts uphill at its maximum; at F = 9.8 N no friction is needed; at the maximum it acts downhill.

Why: Below 9.8 N the ramp block would tend to slide down, so friction acts up. Above it the tendency reverses, so friction acts down.

Uniform rope tension is T = F because the rope and pulleys are ideal.

Why: The ideal free end transmits the applied force through the rope without a tension difference, so T has the same magnitude as F.

The moving pulley and load receive upward force 2T from the two vertical rope segments.

Why: Both vertical rope segments pull up on the moving pulley. Their equal tensions add to 2T, rather than just T.

For upward acceleration a: 2F − mg = ma, so F = m(g+a)/2.

Why: Choose the moving pulley and load as the body. Its upward rope force minus weight must supply ma; solving gives the required free-end pull.

For zero acceleration, F = 10(9.8)/2 = 49 N.

Why: Steady lifting has constant velocity and therefore zero acceleration. Two 49 N rope forces exactly balance the 98 N load weight.

Resolve weight along the slope: Fparallel = mg sin 20°.

Why: Only weight's component parallel to the incline can speed the block along it. The normal force balances the perpendicular component.

Newton’s second law gives a = g sin 20°. The mass cancels.

Why: Set mg sin 20° = ma along the slope. Both the driving weight component and the inertia scale with m, so mass cancels.

N = mg = 49 N, so fₛ,max = μₛN = 19.6 N.

Why: Vertical balance gives N = mg. Multiplying that normal force by μs finds the largest available static-friction magnitude, not its automatic value.

The 12 N push is below that limit. Static friction matches the push, rather than automatically taking its maximum value.

Why: The needed 12 N friction is within the 19.6 N limit. It matches and opposes the push, leaving zero net horizontal force and zero acceleration.

Vertical acceleration is zero: N = mg − F sin 30° = 73 N.

Why: The rope's upward component shares the job of supporting the crate. Zero vertical acceleration gives the remaining support from the floor.

Kinetic friction = 0.2(73) = 14.6 N.

Why: The crate is already sliding, so use kinetic friction with the actual 73 N normal force. It acts against the stated sliding direction.

Horizontal net force = 50 cos 30° − 14.6 = 28.701 N; divide by 10 kg.

Why: The rope's horizontal component pulls forward while friction pulls backward. Divide their net force by the crate's mass to get its acceleration.

Along the slope, mg sin θ − μₖmg cos θ = 0.

Why: Constant speed along a straight slope means zero acceleration. The downhill weight component must equal the uphill kinetic-friction force.

Therefore μₖ = tan θ. The normal force is mg cos θ.

Why: Divide the balance by mg cos θ. This leaves sin θ/cos θ = tan θ, showing why the needed coefficient does not depend on mass.

Choose the 5 kg mass’s downward direction as positive. Summing both equations gives (5 − 3)g = (5 + 3)a.

Why: The rope gives equal acceleration magnitudes in opposite directions. Adding the two mass equations cancels tension and leaves the difference in weights driving both masses.

a = 2.45 m/s². For the 3 kg mass, T − 3g = 3a.

Why: Use the calculated a in the lighter mass's own equation. Its tension must exceed 3g because that mass accelerates upward.

A fixed pulley changes direction without multiplying force: T = mg = 117.6 N.

Why: A fixed pulley only redirects the ideal rope. Constant speed means the single upward tension on the load must balance its weight.

The free end moves 0.5 m, so work = 117.6(0.5).

Why: With a fixed pulley, the free end travels the same distance as the load. The downward pull and downward end displacement are aligned, giving positive work.

Two tensions support the load: 2T = mg.

Why: The moving pulley receives two equal upward tensions. At constant speed those two support forces together balance the load's weight.

Conservation of string length requires the free end to move twice as far as the load.

Why: Raising the pulley shortens both supporting rope segments by 0.3 m. The free end must take up their combined 0.6 m, preserving ideal input work.

The system mass is 6 kg, so a = 12/6 = 2 m/s².

Why: Treat both carts as one system so their internal coupling force cancels. The 12 N external force accelerates the full 6 kg.

The coupling is the only horizontal force on the 4 kg cart, so T = 4a.

Why: Now isolate the 4 kg cart. Its coupling force must provide the ma needed for the same 2 m/s² acceleration.

Acceleration points with net force: a = −6/2 = −3 m/s².

Why: Right-positive makes the leftward net force negative. Dividing it by mass gives a leftward acceleration even while the object initially moves right.

With right positive, v = 5 − 3t. Set v = 0.

Why: The constant net force gives constant acceleration, so use v = v₀ + at. Setting velocity to zero identifies the instant the rightward motion stops.

At terminal speed, acceleration is zero and drag balances weight.

Why: Terminal motion has constant downward speed, so its acceleration is zero. The upward drag must exactly balance the object's downward weight.

0.08v² = 2(9.8), so v = √(19.6/0.08).

Why: Set the stated drag model equal to mg and solve for the positive speed. Direction is downward because this is the falling terminal state.

Vertical balance: T cos 20° = mg.

Why: The rope's angle is measured from vertical, so its upward component is T cos 20°. That component alone balances weight.

Horizontal balance: F = T sin 20° = mg tan 20°.

Why: The horizontal applied force balances the rope's sideways component. Substituting T = mg/cos 20° turns the ratio into tan 20°.

Hooke’s law is Fₓ = −kx.

Why: The spring's restoring force points opposite its signed displacement from the relaxed position. That opposition is the minus sign in Hooke's law.

Fₓ = −80(−0.15) = +12 N. Compression produces a force toward the relaxed position.

Why: Here x is negative because the spring is compressed. The two minus signs give a positive force, pushing the object back toward x = 0.

Both springs have the same extension; their forces add. Thus keff = 100 + 200 = 300 N/m.

Why: Parallel springs stretch by the same x, so their separate kx forces add. Their stiffnesses therefore add to one effective spring constant.

At rest, keff x = mg.

Why: At rest, the two spring forces together balance weight. Divide mg by their combined stiffness to get the common extension.

Each spring carries the same 10 N force, so x₁ = 10/100 = 0.1 m and x₂ = 10/200 = 0.05 m.

Why: In an ideal series arrangement, the same tension passes through both springs. Hooke's law then gives a larger extension for the softer spring.

Total extension is 0.15 m. keff = 10/0.15.

Why: The two extensions add along the line of the pull. The effective constant relates that total extension to the common 10 N force.

Use the dot product W = Fd cos θ.

Why: Work uses the force component along the displacement. The dot product projects this angled pull onto the suitcase's horizontal motion.

W = 25(6) cos 60° = 75 J. The vertical component does no work on a horizontal displacement.

Why: The 60° cosine leaves only the horizontal half of the 25 N force. The vertical force has no vertical displacement on which to do work.

Work is the area under Fₓ versus x: W = ∫₁³4x dx.

Why: Because the force changes with x, one force value cannot represent the whole trip. Add its work contributions as the area under the force–position graph.

W = [2x²]₁³ = 2(9 − 1).

Why: The antiderivative of 4x is 2x². Subtract its value at 1 m from its value at 3 m to count only the stated interval.

Gravity has a component in the downhill direction, so its work is positive.

Why: The displacement is downhill, and gravity's parallel component points downhill too. Their dot product is positive.

Kinetic friction points uphill, opposite motion, so its work is negative.

Why: Sliding friction acts uphill while the displacement is downhill. The force–displacement angle is 180°, giving negative work.

The normal force is perpendicular to the displacement and does zero work.

Why: A stationary incline's normal force stays perpendicular to the block's along-slope displacement. The 90° dot product is zero.

The spring’s potential energy decreases from ½kxi² to ½kxf².

Why: An ideal spring stores ½kx². Moving closer to the relaxed length reduces that stored energy because the final extension is smaller.

Work done by the spring is ½k(xi² − xf²) = 100(0.04 − 0.0025).

Why: Spring work is minus the change in spring potential energy. Subtract the final stored energy from the initial value to find the positive energy delivered.

Their sum is (4, 5) N.

Why: Combine the two known forces component by component to find the force imbalance that the third force must remove.

Equilibrium requires the third force to be the negative of that sum.

Why: Equilibrium requires zero net force in both directions. The missing force therefore has exactly the opposite components to the known sum.

For Hooke’s law, force magnitude divided by extension must be constant.

Why: A single Hooke's-law constant would make F/x the same at every measured extension, with the graph passing through the relaxed origin.

The ratios are 100, 150 and 200 N/m, so the data do not fit a single linear spring constant.

Why: Convert centimetres to metres before comparing F/x. The increasing ratios show that the band's force grows faster than one constant k would predict.