Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — resultant of two forces
Find the magnitude and direction of the resultant of a 5 N force along the x-axis and an 8 N force at 60° to it.
- 1Rx=5+8cos60°=5+4=9Resolve each force, then add components.
- 2Ry=0+8sin60°=6.93The 5 N force has no vertical component.
- 3∣R∣=92+6.932=129=11.4 NNot 5+8=13 — the forces are not parallel.
- 4θ=tan−196.93=37.6° above the x-axisBetween 0° and 60°, as it must be.
Example 2 — a particle in equilibrium under three forces
A body of weight 100 N hangs from two strings at 40° and 55° to the horizontal, on opposite sides. Find the tensions.
- 1Horizontally: T1cos40°=T2cos55°The horizontal components must cancel.
- 2Vertically: T1sin40°+T2sin55°=100Both tensions have upward components.
- 3From the first, T1=T2cos40°cos55°=0.7487T2
- 4Substituting: 0.7487T2sin40°+T2sin55°=100
- 5T2(0.4813+0.8192)=100, so T2=76.9 N
- 6T1=0.7487×76.9=57.6 NThe steeper string carries more — check that it does.
Example 3 — a block on a rough inclined plane
An 8 kg block rests on a plane at 25° with μ=0.3. Show that it slides and find its acceleration. Take g=9.81 m s−2.
- 1Perpendicular to the plane: R=mgcos25°=8(9.81)(0.9063)=71.1 NNot mg — the plane is inclined.
- 2Greatest friction: μR=0.3×71.1=21.3 N
- 3Down the slope: mgsin25°=8(9.81)(0.4226)=33.2 N
- 433.2>21.3, so friction cannot hold it and the block slides.Compare the two before assuming motion.
- 5ma=mgsinα−μmgcosα, so a=g(sin25°−0.3cos25°)The mass cancels.
- 6a=9.81(0.4226−0.2719)=1.48 m s−2 down the slope