Problem : A block of mass “m” is at rest on a rough incline of angle “θ”. Find the contact forces and the net contact force on the block.

Figure 1: The block rests on the incline.

A block on an incline

Solution : The forces on the block are (i) its weight (ii) normal force and (iii) Friction force. These forces are not concurrent (see figure).

Figure 2: Forces are not concurrent.

Forces on the block

However, no turning effect is involved. We can, therefore, treat force system concurrent with the "center of mass" of the block. In order to analyze the forces, we consider a coordinate system as shown in the figure. Since the block is at rest, the forces on the block are balanced :

Figure 3: Forces are shown with block as a point with concurrent forces .

Free body diagram

∑ F x = m g sin θ − F f = 0 ∑ F x = m g sin θ − F f = 0

∑ F y = N − m g cos θ = 0 ∑ F y = N − m g cos θ = 0

There are two contact forces (i) normal force and (ii) friction. The friction is given by the first equation :

F f = m g sin θ F f = m g sin θ

The normal force is given by the second equation :

N = m g cos θ N = m g cos θ

The net contact force is vector sum of two contact forces, which are at right angle to each other. Hence, magnitude of net contact force is :

F C = N 2 + F f 2 F C = N 2 + F f 2

F C = { m g 2 cos 2 θ + sin 2 θ } F C = { m g 2 cos 2 θ + sin 2 θ }

⇒ F C = m g ⇒ F C = m g

Problem : : A body is subjected to four coplanar concurrent forces as shown in the figure. The body is under equilibrium. Find the forces “ F 1 F 1 ” and “ F 2 F 2 ”.

Figure 4: The forces are concurrent.

Four forces on a body

Solution : The body is in equilibrium under the action of coplanar concurrent forces. It means that forces form a balanced force system. We can treat the body as a point on the free body diagram. Let “x” and “y” be the axes of the coplanar coordinate system as shown in the figure. Then,

Figure 5: The forces are concurrent.

Free body diagram

∑ F x = 16 + 8 sin 30 0 − F 2 cos 30 0 = 0 ∑ F x = 16 + 8 sin 30 0 − F 2 cos 30 0 = 0

∑ F y = F 1 + 8 cos 30 0 − F 2 sin 30 0 = 0 ∑ F y = F 1 + 8 cos 30 0 − F 2 sin 30 0 = 0

We can find “ F 2 F 2 ” from the first equation,

⇒ F 2 X 3 2 = 16 + 8 X 1 2 = 20 ⇒ F 2 X 3 2 = 16 + 8 X 1 2 = 20

⇒ F 2 = 20 X 2 3 = 40 3 N ⇒ F 2 = 20 X 2 3 = 40 3 N

Substituting in second equation and solving for “ F 1 F 1 ”, we have :

⇒ F 1 = F 2 sin 30 0 − 8 cos 30 0 = 40 3 X 1 2 − 8 X 3 2 ⇒ F 1 = F 2 sin 30 0 − 8 cos 30 0 = 40 3 X 1 2 − 8 X 3 2

⇒ F 1 = 20 3 − 4 X 3 ⇒ F 1 = 20 3 − 4 X 3

⇒ F 1 = 20 − 12 3 = 8 3 N ⇒ F 1 = 20 − 12 3 = 8 3 N

Problem : A spherical mass “m” hanging from ceiling is displaced by applying a horizontal force,F. The string makes an angle “θ” with vertical as shown in the figure. Find force F.

Figure 7: The forces are concurrent.

Three forces on a spherical mass

Solution : Here, we shall employ Lami’s theorem as there are exactly three forces acting on the spherical mass. The FBD of the spherical mass is as shown in the figure.

Figure 8: The forces are concurrent.

Free body diagram

Applying Lami’s theorem, we have :

m g sin 90 0 + θ = T sin 90 0 = F sin 180 0 − θ m g sin 90 0 + θ = T sin 90 0 = F sin 180 0 − θ

⇒ m g cos θ = T 1 = F sin θ ⇒ m g cos θ = T 1 = F sin θ

From the last two ratios, we have :

⇒ F = T sin θ ⇒ F = T sin θ

From the first two ratios, we have :

⇒ T = m g cos θ ⇒ T = m g cos θ

Combining two equations,

⇒ F = m g sin θ cos θ = m g tan θ ⇒ F = m g sin θ cos θ = m g tan θ

Lami's theroem provides an alternative to normal component analysis - may be with some computational advantage. Nevertheless, we should keep in mind that it is just an alternative method. We could have solved this problem, using component analysis as :

∑ F x = F - T sin θ = 0 ⇒ F = T sin θ ∑ F x = F - T sin θ = 0 ⇒ F = T sin θ

and

∑ F y = mg - T cos θ = 0 ⇒ mg = T cos θ ∑ F y = mg - T cos θ = 0 ⇒ mg = T cos θ

Taking ratio to eliminate “T”,

⇒ F mg = tan θ ⇒ F = mg tan θ ⇒ F mg = tan θ ⇒ F = mg tan θ