Turning Effect of Forces

1.Find the resultant of the following forcers

(i)                 10N along x-axis

(ii)               6N along y-axis

(iii)             4N along negative x-axis

Given Data

F_x1 = 10N   along   +ve    x-axis

F_x2 = 4N   along   -ve    x-axis

Fy = 6N   along       y-axis

F_x1 = F_x1 - F_x2 = 10 - 4  = 6 N  (The Resultant force along X-axis)

To Find

    ϴ=?F =?

Formula

Calculation

2.Find the perpendicular components of a force of 50N making an angle 30⁰ with x-axis.

Given Data

     ϴ=30^oF= 50N

To Find

                F_x = ?   

                F_y = ?   

Formula

               F_x = FCosϴ   

               F_y = FSinϴ  

Calculation

As

            F_x = FCosϴ   

Putting the values

            F_x = 50 Cos30 = 50 × 0.866 =43.3N   

Similarly

            F_y = FSinϴ  

Putting the values

            F_y = 50 Sin30 = 50 × 0.5  = 25 N  

Hence horizontal force will be 43.3N and vertical force will be 25N.

3.Find the magnitude and direction of a force if its x-component is 

12N and y-component is 5N.

Given Data

F_x = 12 N   

 F_y = 5   

To Find

F = ?

ϴ = ?

Formula

Calculation

4.A force of 100N is applied perpendicularly on a spanner at a distance 

of 10cm from aunt. Find the torque produced by the force?

Given Data

                     F = 100N

                     r = 10cm = 10/100 =0.1m

To Find

                     t = ?

Formula

                     t =  F× r

Calculation

as

                     t =  F× r

Putting the values

                     t  = 100 ×0.1 = 10Nm

Hence the torque will be 10Nm.

5.A force is acting on a body making angle of 30⁰ with the horizontal. 

The horizontal component of the force is 20N. Find the force?

Given Data

                     F_x = 20 N   

                     ϴ = 30

To Find

                     F = ?

Formula

                   F_x = FCosϴ   or  F = F_x / Cosϴ   

Calculation

                  as

                                      F = F_x /Cosϴ   

                Putting the values

                                     F = 20/Cos30 = 20/0.866

                                      F = 23.1 N

Hence the applied force will be 23.1N.

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

Given Data

                     F = 150N

                     r = 16cm = 16/100 =0.16m

To Find

                     t = ?

Formula

                     t =  F× r

Calculation

as

              rorque due to couple = t = 2 F× r

Putting the values

                     t  = 150 ×0.16 = 16Nm

Hence the torque due to couple will be 16Nm.

7.A pieture frame is hanging by two vertical strings. The tensions in the string are 3.8N and 4.4N. Find the weight of the picture frame?

Given Data

                  F_y1 = 3.8N   

                 F_y2 = 4.4N   

To Find

               w = ?

Formula

According to first condition of equilibrium

              Sum of all forces  = 0

Therefore

Upwards forces = downwards forces

              F_y1 + F_y1 = w

Calculation

as

              F_y1 + F_y1 = w

Therefore

             3.8 + 4.4 = 8.2N = w

             w = 8.2 N

hence the weight of the picture will be 8.2N

8.Two blocks of 5kg and 3kg are suspended by the two strings 
find the tension in each string?

Given Data

                   m₁  = 5kg

                   m₁  = 3kg

To Find

                    Tension due to body 1. T₁ = ?

                    Tension due to body 2.T₂ = ?

                   (whole tension due to both boodies)T₂^’ = ?

Formula

                  T₁ = m₁g

                  T₂ = m₂g

                  T₂^’ = T₁ + T₂

Calculation

               as

                   T₁ = m₁g

              Putting the values

                  T₁ = 3 × 10 = 30N

                as

                  T₂ = m₂g

              Putting the values

                  T₂ = 5 × 10 = 50N

                as

                 T₂^’ = T₁ + T₂

              Putting the values

                T₂^’ = = 30 + 50 =80N

Hence the lower body face 30N tension while upper body face 80N tension!

9.a nut has been tightened by a force of 200N using 10cm long spanner 

what length of spanner is required to loosen the same nut with 150 force?

Given Data

                          F₁ = 200N

                          r₁ = 10cm = 10/100 = 0.1m

                          F₂ = 150N

To Find

                           r₂ = ?

Formula

                              According to the 2n condition of equilibrium

                              F₁ ×  r₁ = F₂ ×  r₂

Calculation

                           As

                            F₁ ×  r₁ = F₂ ×  r₂

                  Putting the values

                  

                            200 × 0.1 = 150 × r₂

                                 20/150 = r₂

                                0.133m = r₂

                                        r₂ = 0.133 × 100 cm = 13.3cm

Hence the length of spanner must be 13.3cm.

10.A block of mass 10kg is suspended at a distance of 20cm 

from the center of a uniform bar 10 long. What force is required to balance

 it at its center of gravity by applying the force at the other end of the bar?

Given Data

                         m = 10 kg

                         w =F₁­ = mg = 10×10 = 100N

                        distance from center = r₁ = 20cm =0.2m

                        distance from end to center = r₂ =50cm =0.5m

To Find

                      F₂ = ?

Formula

                     According to the 2n condition of equilibrium

                      

                     F₁ ×  r₁ = F₂ ×  r₂

Calculation

                     As

                     F₁ ×  r₁ = F₂ ×  r₂

              Putting the values

                      100 ×0.2  = F₂ × 0.5

                            20/0.5 = F₂

                               40N = F₂

                                           F₂ = 40N

Hence the we need 40N of force!