9th Class Physics Chapter 4 (Numerical Problems with Solutions)

UNIT No. 4

Turning Effect of Forces

Numerical Problems

4.1         Find the resultant of the following forces:

(i)            10 N along x-axis

(ii)         6 N along y-axis and

(iii)       4 N along negative x-axis.

4.2         Find the perpendicular components of a force of 50 N making an angle of 30° with x axis.

Data

Force = F = 50 N

Angle = Ѳ = 30^o

x-component of force = F_x = ?

y-component of force = F_y = ?

Formula

F_x = FcosѲ

F_y = FsinѲ

Solution

x-component of force

F_x = FcosѲ

F_x = (50)cos(30)

F_x = 43.3 N        Ans

y-component of force

F_y = FsinѲ

F_y = (50)sin(30^o)

F_y = 25 N           Ans

4.3         Find the magnitude and direction of a force, if its x-component is 12 N and y- component is 5 N.

Data

x-component of force = Fx = 12 N

y-component of force = Fy = 5 N

magnitude of force = F = ?

direction of force = Ѳ = ?

Formula

$$F = \sqrt{F_{x}^{2}+F_{y}^{2}}$$

$$\Theta = tan^{-1}(\frac{F_{y}}{F_{x}})$$

Solution

magnitude of force

$$F = \sqrt{F_{x}^{2}+F_{y}^{2}}$$

$$F = \sqrt{(12)^{2}+(5)^{2}}$$

$$F = \sqrt{144 + 25}$$

$$F = \sqrt{169}$$

F = 13 N            Ans

direction of force

$$\Theta = tan^{-1}(\frac{F_{y}}{F_{x}})$$

$$\Theta = tan^{-1}(\frac{5}{14})$$

$$\Theta = tan^{-1}(0.416)$$

Ѳ = 22.6^o with x-axis    Ans

 

4.4         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.

Data

force = F = 100 N

moment arm of force = L = 0.1 m

torque produced by the force. = τ = ?

Formula

τ = FL

Solution

τ = FL

τ = (100 N)(0.1 m)

τ = 10 N            Ans

 

4.5         A force is acting on a body making an angle of 30° with the horizontal. The horizontal component of the force is 20 N. Find the force,

Data

Angle = Ѳ = 30^o

horizontal component of force = F_x = 20 N

Force = F = ?

Formula

$$F_{x}=Fcos\Theta$$

$$F = \frac{F_{x}}{Fcos\Theta }$$

 Solution

$$F_{x}=Fcos\Theta$$

$$F = \frac{F_{x}}{Fcos\Theta }$$

$$F = \frac{20}{Fcos30 }$$

$$F = \frac{20}{0.866}$$

F = 23.1 N         Ans

 

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

Data

Radius of steering of a car = r = 16 cm = 0.16 m

force = F = 50 N

torque produced by the couple. = τ = ?

Formula

The

torque of a couple =  (Magnitude of one of the two forces)(perpendicular distance between them)

torque of a couple =  (F)(2r)

 

Solution

torque of a couple =  (F)(2r)

torque of a couple =  (50 N)(2x0.16m)

torque of a couple =  16 Nm   Ans

 

4.7         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.

Data

Tension in the first string = T₁ = 3.8 N

Tension in the second string = T₂ = 4.4 N

Weight of the picture frame = w = ?

Formula

ΣF_y = 0

Solution

ΣF_y = 0

T_{1 +} T₂ - w = 0

T_{1 +} T₂  = w

w = T_{1 +} T₂  

w = 3.8 N + 4.4 N

w = 8.2 N          Ans

4.8         Two blocks of masses 5 kg and 3 kg are suspended by the two strings as shown. Find the tension in each string.

Data

Mass of 1^st block = m₁ = 5 kg

Mass of 2^nd block = m₂ = 3 kg

weight of 1^st block = w₁ = m₁g = (5)(10) = 50 N

weight of 2^nd block = w₂ = m₂g =  (3)(10) = 30 N

Tension in the 1^st string = T₁ = ?

Tension in the 2^nd string = T₂ = ?

Formula

ΣF_y = 0

Solution

Tension in the 1^st string

ΣF_Y1 = 0

T₁  - w₁ - w₂ = 0

T₁  = w₁ + w₂

T₁  = 50 N + 30 N

T₁  = 80 N          Ans

Tension in the 2^nd string

ΣF_Y2 = 0

T₂  - w₂ = 0

T₂  =  w₂

T₁  =  30 N

T₂  = 30 N          Ans

 

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

Data

1^st force = F₁ = 200 N

Length of spanner using 1^st force = L₁ = 0.1 cm

2^nd force. = F₂ = 150 N

Length of spanner for using 2^nd force = L₂ = ?

Formula

τ = FL

Solution

To loosen the same nut same torque is required from both forces i.e.

Torque produced by F₁ = Torque produced by F₂

τ₁ = τ₂

F₁L₁ = F₂L₂

L₂ = 0.133 m     Ans

 

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

First, we will draw a diagram for this problem

 

Data

Mass of block = m 10 kg

Weight of the block = w = mg = (10)(10) = 100 N

Distance AG = 20 cm = 0.2 m  (Given in problem)

Distance GB = 50 cm = 0.5 m   (From diagram)

force required to balance the rod at its centre of gravity = F =?

Formula

According to principle of moments

Clockwise moments = Anticlockwise moments

Solution

Applying principle of moments on this problem

Clockwise moments = Anticlockwise moments

w  x AG  =  F x GB

$$F = \frac{(w)(AG)}{GB}$$

$$F = \frac{(100)(0.2)}{0.5}$$

           F = 40 N             Ans