Numerical Problems

Oct 12, 2022

Find the resultant of the following forces: 10 N along the x-axis 6 N along the y-axis and 4 N along the negative x-axis. (8.5 N making 45 ⁰ with x-axis)

Difficulty: Medium

Solution: $F_{x}$ = Net force along x-axis = 10.4 = 6 N $F_{y}$ = Force along y-axis = 5 N The magnitude of the resultant force = F =?   F = $√F_{x}^{2} + F_{y}^{2}$ F = $√(6)^{2} + (6)^{2}$ F = √36 + 36    = √72 = 8.5 N   Now, θ = $tan^{-1}$ =$\frac{F_{y}}{F_{x}}$                 θ = $tan^{-1}$ =$\frac{6}{6}$  θ = $tan^{-1}$ (1) θ = 45⁰ with x-axis

Find the perpendicular components of a force of 50 N making an angle of 30⁰ with an x-axis. (43.3 N, 25 N)

Difficulty: Easy

Solution: Force F = 50 N Angle θ = 30⁰ $F_{x}$ = ? and $F_{y}$ = ? $F_{x}$ = F cos θ $F_{x}$= $50 \timescos 30$       = $50 N\times 0.866$ (˙.˙ cos 30⁰ = 0.866) $F_{x}$ = 43.3 N   Similarly, $F_{y}$ = F sin θ $F_{y}$= $50 \times 0.5$(˙.˙ sin 30⁰ = 0.5) $F_{y}$= 25 N

Find the magnitude and direction of a force, if its x-component is 12 N and y-component is 5 N. (13 N making 22.6⁰ with x-axis)

Difficulty: Medium

Solution: $F_{x}$ = 12 N   $F_{y}$= 5 N   Magnitude of the force = F = ? Direction of the force = θ = ?   F = $√F_{x}^{2} + F_{y}^{2}$   F = $√(12)^{2} + (5)^{2}$   F = √144+25   F = 13 N     θ = $tan^{-1} =\frac{F_{y}}{F_{x}}$   θ = $tan^{-1} =\frac{12}{5}$    θ = $tan^{-1}$ (2.4)   θ = 22.6⁰ with x-axis