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

A force of 100 N is applied perpendicularly on a spanner at a distance of 10 cm from a nut. Find the torque produced by the force. (10 Nm)

Difficulty: Easy

Solution: Force = F = 100 N        Distance = L = 10 cm = 0.1 m        Torque Τ =? Torque Τ = $F \times L$                 = 100 $N \times 0.1 m$                 = 10 Nm

A force is acting on a body making an angle of 30⁰ with the horizontal. The horizontal component of the force is 20 N. Find the force. (23.1 N)

Difficulty: Easy

Solution: Angle θ = 30⁰ (with x-axis) Horizontal component of force $F_{x}$ = 20 N Force F = ? $F_{x}$ = F cos θ   20 N = F cos 30⁰ 20 N = $F \times 0.866$ (˙.˙ cos 30⁰ = 0.866) F =$\frac{ 20 N}{0.866}$= 23.09 F = 23.1 N

The steering of a car has a radius of 16 cm. Find the torque produced by a couple of 50 N. (16 Nm)

Difficulty: Easy

Solution: Radius = r = L = 16 cm =$\frac{16}{100}$ m = 0.16 m        Couple arm = L = 16 cm =$\frac{16}{100 }$ m = 0.16 m        Force = F = 50 N        Torque Τ = ? Torque Τ = $F \times L$                 = $50 N\times(2 \times 0.16)$                 = 16 Nm

A picture frame is hanging by two vertical strings. The tensions in the strings are 3.8 N and 4.4 N. Find the weight of the picture frame. (8.2 N)

Difficulty: Easy

Solution: Tension $T_{1}$ = 3.8 N Tension $T_{2}$ = 4.4 N Weight of the picture frame = w =?   When the picture is in equilibrium, then   ∑ $F_{x}$ = 0 and ∑ $F_{y}$ = 0 Therefore T – w = 0   Or $(T_{1} + T_{2})$ – w = 0               $T_{1} + T_{2}$ = w           3.8 + 4.4 = w                     W = 8.2 N

Two blocks of mass 5 kg and 3 kg are suspended by the two strings as shown. Find the tension in each string. (80 N, 30 N)

Difficulty: Medium

Solution: Mass of large block = M = 5 kg        Mass of large block = m = 3 kg        Tension produced in each string = $T_{1} =  ?  and  T_{2}$ = ?                    $T_{1}$ = $w_{1} + w_{2}$             $T_{1}$ = Mg + mg            $T_{1}$ = (M + m)g            $T_{1}$ = $(3+5) \times 10$                = $8 \times 10$                = 80 N                        Also, $T_{2}$ = mg            $T_{2} = 3 \times10$ = 30 N

A nut has been tightened by a force of 200 N using a 10 cm long spanner. What length of a spanner is required to loosen the same nut with 150 N force? (13.3 cm)

Difficulty: Medium

Solution: Force = $F_{1}$ = 200 N Length = $L_{1}$ = 10 cm = $\frac{10 }{100}$= 0.1 m          Length of the spanner to tighten the same nut: Force = $F_{2}$ = 150 N Length = $L_{2}$ = ?          Since $Τ_{1} = Τ_{2}$     $F_{1} \times L_{1}$ =  $F_{2} \times L_{2}$ $200 \times 0.1$ = $150 \times L_{2}$ 20 = $150 \times L_{2}$   $L_{2}$ = $\frac{20}{150}$ = 0.133 m = $0.133 \times 100$ = 13.3 cm

A block of mass 10 kg is suspended at a distance of 20 cm from the center of a uniform bar 1 m long. What force is required to balance it at its center of gravity by applying the force at the other end of the bar? (40 N)

Difficulty: Medium

Solution: Mass of the block = m = 10 kg        Length of the bar = l = 1 m        Moment arm of $w_{1} = L_{1}$ = 20 cm = 0.2 mMoment arm of $w_{2} = L_{2}$ = 50 cm = 0.5 m    Did force require to balance the bar $F_{2}$ =?    By applying principle of moments:     Clockwise moments = anticlockwise moments    $F_{1} \times L_{1}$ =  $F_{2} \times L_{2}$    $mg \times L_{1}$ =  $F_{2} \times L_{2}$    $(10 \times10 )\times 0.2$ =  $F_{2}\times0.5$    20 = $F_{2}\times 0.5$    $F_{2} =\frac{20}{0.5}$ = $\frac{200}{5}$= 40 N

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