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.

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 :

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

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

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

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

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

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

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

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

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

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

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

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

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