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Resolution of forces (application)

Module by: Sunil Kumar Singh. E-mail the author

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Hints for solving problems

We resolve a force along the axes of a coordinate system (see Components of a vector ) in following manner :

1 : Select coordinate system such that maximum numbers of forces are along the axes of chosen coordinate system.

2 : Determine “x” and “y” components of force by considering acute angle between the direction of axis and force.

3 : Use cosine of the acute angle for the component of axis with which angle is measured. Use sine of the angle for the axis perpendicular to the other axis. If “θ” be the angle that the force vector makes with x – axis, then components along “x” and “y” axes are :

F x = F cos θ F x = F cos θ

F y = F sin θ F y = F sin θ

4 : If the component i.e. projection of force is in the opposite direction of the reference axis, then we prefix the component with a negative sign.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the analysis framework for law of motion. The questions are categorized in terms of the characterizing features of the subject matter :

  • Balanced forces
  • Unbalanced forces
  • Acceleration due to gravity
  • Forces on an incline

Balanced forces

Problem 1 : Two small spherical objects of mass “m” and “M” are attached with strings as shown in the figure. Find the angle “θ” such that the given system is in equilibrium.

Figure 1: Two blocks are hanging with the help of three strings.
Balanced forces
 Balanced forces  (rfq1.gif)

Solution : Every point of the system is under action of balanced force system. We shall work through the points "A" and "B", where masses are attached.

Figure 2: Forces on the spherical objects are weights and tensions as shown.
Forces on the objects
 Forces on the objects  (rfq2a.gif)

Let “ T 1 T 1 ” and “ T 2 T 2 ” be the tensions in the two strings, meeting at point "A". The forces at "A" are shown in the figure above. Considering balancing of forces in "x" and "y" directions :

F x = T 1 cos 45 0 - T 2 cos 45 0 = 0 F x = T 1 cos 45 0 - T 2 cos 45 0 = 0

T 1 = T 2 = T say T 1 = T 2 = T say

F y = T 1 sin 45 0 + T 1 sin 45 0 = m g F y = T 1 sin 45 0 + T 1 sin 45 0 = m g

T sin 45 0 + T sin 45 0 = m g T sin 45 0 + T sin 45 0 = m g

2 T 2 = m g 2 T 2 = m g

T = m g 2 T = m g 2

Let “ T 3 T 3 ” be the tension in the upper section. The forces on the mass “M” is shown in the figure above. Considering balancing of forces in "x" and "y" directions :

F x = T 3 cos θ - T sin 45 0 = 0 F x = T 3 cos θ - T sin 45 0 = 0

Substituting for “T” and evaluating, we have :

T 3 cos θ = m g 2 X 1 2 = m g 2 T 3 cos θ = m g 2 X 1 2 = m g 2

F y T 3 sin θ = T cos 45 0 + M g F y T 3 sin θ = T cos 45 0 + M g

Substituting for "T",

T 3 sin θ = T 2 + M g = m g 2 X 1 2 + M g T 3 sin θ = T 2 + M g = m g 2 X 1 2 + M g

T 3 sin θ = mg 2 + M g T 3 sin θ = mg 2 + M g

Taking ratio of the resulting equations in the analysis of forces on "M",

tan θ = m g 2 + M g m g 2 tan θ = m g 2 + M g m g 2

tan θ = 1 + 2 M m tan θ = 1 + 2 M m

θ = tan 1 1 + 2 M m θ = tan 1 1 + 2 M m

Unbalanced forces

Problem 2 : A force “F” produces an acceleration “a” when applied to a body of mass “m”. Three coplanar forces of the same magnitudes are applied on the same body simultaneously as shown in the figure. Find the acceleration of the body.

Figure 3: Three forces act on a body as shown.
Forces on a body
 Forces on a body  (rfq10.gif)

Solution : We select two axes of coordinate system so that they align with two mutually perpendicular forces as shown in the figure. We keep in mind that three forces are coplanar.

Figure 4: Coordinate system is selected to align with force.
Coordinate system
 Coordinate system  (rfq9.gif)

Taking components of forces in “x” and “y” directions,

F x = F F cos 30 0 = F 1 3 2 = F 2 3 2 F x = F F cos 30 0 = F 1 3 2 = F 2 3 2

F y = F F sin 30 0 = F 1 1 2 = F 2 F y = F F sin 30 0 = F 1 1 2 = F 2

The net force is given by :

F n e t = F x 2 + F y 2 = F 2 { 1 + 2 3 2 } F n e t = F x 2 + F y 2 = F 2 { 1 + 2 3 2 }

F n e t = F 2 1 + 4 + 3 4 3 F n e t = F 2 1 + 4 + 3 4 3

F net = F 2 8 4 3 F net = F 2 8 4 3

F net = F 2 3 F net = F 2 3

But, it is given that F = ma. Substituting for "F" in the equation, we have :

F net = m a 2 3 F net = m a 2 3

Let acceleration of the body under three coplanar forces be a’. Then,

m a = m a 2 3 m a = m a 2 3

a = a 2 3 a = a 2 3

Acceleration due to gravity

Problem 3 : A small cylindrical object slides down the smooth groove of 10 m on the surface of an incline plane as shown in the figure. If the object is released from the top end of the groove, then find the time taken to travel down the length.

Figure 5: The cylindrical object slides down the smooth groove.
Motion along a groove
 Motion along a groove  (rfq3a.gif)

Solution : In order to find the time taken to travel down the incline, we need to know acceleration along the groove. The object travels under the influence of gravity. The component of acceleration due to gravity along the incline (GE), as shown in the figure below, is :

a = g cos 60 0 a = g cos 60 0

Figure 6: Acceleration due to gravity is resolved along the groove in two stages.
Component of acceleration due to gravity
 Component of acceleration due to gravity  (rfq4a.gif)

This component of acceleration makes an angle 60° with the groove. Hence, component of acceleration along the groove (CA) is given as :

a = a cos 60 0 a = a cos 60 0

Combining two equations, the acceleration along the groove, "a", is :

a = g cos 60 0 cos 60 0 = g X 1 2 X 1 2 = g 4 a = g cos 60 0 cos 60 0 = g X 1 2 X 1 2 = g 4

The acceleration along the groove is constant. As such, we can apply equation of motion for constant acceleration :

x = u t + 1 2 a t 2 x = u t + 1 2 a t 2

10 = 0 + 1 2 X g 4 X t 2 10 = 0 + 1 2 X g 4 X t 2

t = ( 80 g ) = 8 = 2 2 s t = ( 80 g ) = 8 = 2 2 s

Forces on an incline

Problem 4 : Four forces act on a block of mass “m”, placed on an incline as shown in the figure. Then :

Figure 7: A force "F" acts as shown.
A block on an incline
 A block on an incline  (rfq5.gif)

  1. Resolve forces along parallel and perpendicular to incline and find net component forces in two directions.
  2. Resolve forces along horizontal and vertical directions and find net component forces in two directions.

Solution : The free body diagram with four forces with coordinates are as shown in the figure below.

Figure 8: Forces are resolved in directions parallel and perpendicular to incline.
A block on an incline
 A block on an incline  (rfq6.gif)

The net component of forces along two axes are :

F x = F cos α F F m g sin θ F x = F cos α F F m g sin θ

F y = N + F sin α m g cos θ F y = N + F sin α m g cos θ

Now, we select “x” and “y” axes along horizontal and vertical directions as shown. Note the angles that different forces make with the axes.

Figure 9: Forces are resolved in horizontal and vertical directions.
A block on an incline
 A block on an incline  (rfq7.gif)

The net component of forces along two axes are :

F x = F cos α + θ N sin θ F F cos θ F x = F cos α + θ N sin θ F F cos θ

F y = N cos θ + F sin α + θ m g F F sin θ F y = N cos θ + F sin α + θ m g F F sin θ

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