We have selected three examples in this category. First two examples are analysis of balanced force of typical “string and block” system with increasing complexity. Second example, however, is different in that the string is not “mass-less” as generally considered.

Problem 1 : 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 1

Balanced force system

Solution : Let us consider the spherical mass as the body system.

Free body diagram of spherical mass Free body diagram of spherical mass

The external forces are (i) weight of sphere “mg” (ii) Tension, T, in the string and (iii) force, F, applied in horizontal direction.

Figure 2

Balanced force system

∑ 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

Problem 2 : A block weighing 100 N is suspended with the help of three strings as shown in the figure. Find the tension in each of the string.

Figure 3

Balanced force system

Solution : This example illustrates one important aspect of force diagram. We can even draw force diagram of a point on the system like “O”, where three strings meet. The point does not represent a body, but force diagram is valid so long we display the forces acting through the point, O.

Let T 1 T 1 , T 2 T 2 and T 3 T 3 be the tensions in the string as shown in the figure here.

Figure 4

Balanced force system

A preliminary assessment of forces suggests that analysis of forces on block will provide value for the unknown. Hence, we first analyze force on the block.

Free body diagram of body Free body diagram of body

T 1 = 100 N T 1 = 100 N

Figure 5

Balanced force system

Free body diagram of “O” Free body diagram of “O”

The external forces at point “O” are (i) Tension, T 1 T 1 (ii) Tension, T 2 T 2 and (iii) Tension, T 3 T 3

∑ F x = T 3 sin 60 0 - T 2 = 0 ⇒ T 2 = T 3 sin 60 0 ∑ F x = T 3 sin 60 0 - T 2 = 0 ⇒ T 2 = T 3 sin 60 0

and

∑ F y = T 3 cos 60 0 - T 1 = 0 ⇒ T 3 = T 1 cos 60 0 = 200 N ∑ F y = T 3 cos 60 0 - T 1 = 0 ⇒ T 3 = T 1 cos 60 0 = 200 N

Putting this in the equation for T 2 T 2 , we have :

⇒ T 2 = T 3 sin 60 0 = 200 x √ 3 2 = 100 √ 3 N ⇒ T 2 = T 3 sin 60 0 = 200 x √ 3 2 = 100 √ 3 N

We should note that direction of tension T1 acts up with respect to the body, whereas T1 acts down with respect to point “O”. We need not be overly concerned and just try to figure out, what a taut string does to the body or point in consideration. The tension pulls down the point “O” and pulls up the body. For this reason, it has different directions with respect to them.

Problem 3 : A body of mass “M” is hanging with a string having linear mass density “k”. What is the tension at point “A” as shown in the figure.

Figure 6

Balanced force system

Solution : Here, we must understand that tension in the string is not same. The tension above “A” balances the weight of the block and the weight of the string below point “A”.

Free body diagram of point “A” Free body diagram of point “A”

m s = k ( T - y ) m s = k ( T - y )

Mass of the string below “A”,

The external forces at point “A” are (i) Tension, T (ii) weight of block, mg, and (iii) weight of string below “A”, m s g m s g .

Figure 7

Balanced force system

∑ F y = T - mg - m s g = 0 ⇒ T = ( m + m s ) g ⇒ T = { m + k ( L - y ) } g ∑ F y = T - mg - m s g = 0 ⇒ T = ( m + m s ) g ⇒ T = { m + k ( L - y ) } g

Problem 1 : Find the force, F, required to keep the block stationary on an incline, which makes an angle θ with base as shown in the figure.

Figure 8

Balanced force system

Solution : We can either have (a) axes in horizontal and vertical directions or (b) along incline and perpendicular to incline. Which of the two is better suited here ? We find that both coordinate systems are equally suited. Hence, we can use either of coordinate systems.

Free body diagram of the block Free body diagram of the block

The external forces at point “O” are (i) Force, F (ii) Weight, mg and (iii) Normal force, N.

Figure 9

Balanced force system

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

and

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

Taking ratio, we have :

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

Problem 1 : A string going over a pulley “A” of mass “m” supports a mass “M” as shown in the figure. Find the magnitude of force exerted by the clamp “B” on pulley “A”.

Figure 10

Balanced force system

Solution : Here, we consider pulley as the body system. Let us also consider that clamp “B” exerts a force “F” in an arbitrary direction, making an angle with the horizontal.

We should note that pulley, unless otherwise specified, is considered to be of negligible mass and friction-less. In this case, however, pulley has finite mass “m” and its weight should be considered to be an external force on the pulley.

Free body diagram of pulley Free body diagram of pulley

The string is single piece. Hence, tension in the string all through out is same. Now, the external forces on pulley are (i) Weight, Mg, of the block (ii) Weight, mg, of the pulley (iii) Tension, T and (iv) force, F applied by clamp “B”.

Figure 11

Balanced force system

∑ F x = F x - Mg = 0 ⇒ F x = Mg ∑ F x = F x - Mg = 0 ⇒ F x = Mg

and

∑ F y = F y - mg - T = F y - mg - Mg = 0 ⇒ F y = ( M + m ) g ∑ F y = F y - mg - T = F y - mg - Mg = 0 ⇒ F y = ( M + m ) g

The force exerted by the clamp,F, is :

F = √ ( F x 2 + F y 2 ( = √ { ( Mg ) 2 + ( M + m ) 2 g 2 } ⇒ F = √ { M 2 + ( M + m ) 2 } F = √ ( F x 2 + F y 2 ( = √ { ( Mg ) 2 + ( M + m ) 2 g 2 } ⇒ F = √ { M 2 + ( M + m ) 2 }

Problem 1 : Two blocks weighing 20 N and 10 N at equilibrium are in contact with each other. Find the force F and the normal reactions between all contact surfaces.

Figure 12

Balanced force system

Solution : The question demands that we draw free body diagram of each of the block separately as we are required to know normal reactions at all surfaces. Here, there are three contact surfaces between (i) A and horizontal surface (ii) B and horizontal surface and (iii) A and B.

A preliminary assessment of forces on the blocks suggests that analysis of forces on B will provide values of unknown force(s). It is so because the forces on B are mutually perpendicular (thus, they would not need to be resolved), if appropriate coordinate system is chosen. Hence, we first analyze force on block B.

Free body diagram of B Free body diagram of B

The external forces are (i) weight of B = 10 N (ii) Normal force applied by A i.e. N 1 N 1 (say) (iii) Normal force applied by surface i.e. N 2 N 2 and (iv) force of 20 N

Figure 13

Balanced force system

∑ F x = N 1 - 20 = 0 ⇒ N 1 = 20 N ∑ F x = N 1 - 20 = 0 ⇒ N 1 = 20 N

and

∑ F y = N 2 - 10 = 0 ⇒ N 2 = 10 N ∑ F y = N 2 - 10 = 0 ⇒ N 2 = 10 N

Free body diagram of B Free body diagram of B

The external forces are (i) weight of A, 20 N (ii) Normal force applied by B i.e. N 3 N 3 (say) = N 1 N 1 = 20 N (iii) Normal force applied by surface i.e. N 4 N 4 and (iv) force, F = ?

Figure 14

Balanced force system

∑ F x = F cos 30 0 - N 3 = 0 ⇒ N 3 = F cos 30 0 ⇒ F = N 3 cos 30 0 = 20 √ 3 2 = 20 √ 3 N ∑ F x = F cos 30 0 - N 3 = 0 ⇒ N 3 = F cos 30 0 ⇒ F = N 3 cos 30 0 = 20 √ 3 2 = 20 √ 3 N

and

∑ F y = N 4 - F 1 sin 30 0 - 20 = 0 ⇒ N 3 = F cos 30 0 ⇒ N 4 = F 1 sin 30 0 + 20 = 10 √ 3 x 1 2 + 20 = ( 20 + 5 √ 3 ) N ∑ F y = N 4 - F 1 sin 30 0 - 20 = 0 ⇒ N 3 = F cos 30 0 ⇒ N 4 = F 1 sin 30 0 + 20 = 10 √ 3 x 1 2 + 20 = ( 20 + 5 √ 3 ) N

Problem 1 : The blocks A and B weighing 10 N and 20 N are connected by a string. The block B, in turn, is connected to block C with another string passing over a pulley. Friction forces at all interfaces is negligible. Find the weight of C and tensions in the two strings.

Figure 15

Balanced force system

Solution : There are two strings. Hence, the tensions in the strings may be different. Let T 1 T 1 and T 2 T 2 be the tension in strings AB and BC respectively.

Looking at the various body systems, we guess that the simplest force system is the one associated with block B. However, force analysis of block C will not yield anything as we do not know its weight or the tension T 2 T 2 .

Thus, we begin with block A.

Free body diagram of A Free body diagram of A

Figure 16

Balanced force system

The external forces are (i) weight of A = 10 N (ii) Normal force applied by incline i.e. N 1 N 1 (say) (iii) tension in AB, T 1 T 1 .

∑ F x = 10 sin 30 0 - T 1 = 0 ⇒ T 1 = 10 sin 30 0 = 5 N ∑ F x = 10 sin 30 0 - T 1 = 0 ⇒ T 1 = 10 sin 30 0 = 5 N

We need analyze forces in y – direction as we are not required to determine normal force N 1 N 1 and is not expected to used for analyzing force on block B.

Free body diagram of B Free body diagram of B

The external forces are (i) weight of B, W C W C = 20 N (ii) Normal force applied by incline i.e. N 2 N 2 (say) (iii) tension in AB, T 1 T 1 and (iv) tension in BC, T 2 T 2 .

∑ F x = 20 sin 30 0 + T 1 - T 2 = 0 ⇒ T 2 = 20 x 1 2 + 5 = 15 N ∑ F x = 20 sin 30 0 + T 1 - T 2 = 0 ⇒ T 2 = 20 x 1 2 + 5 = 15 N

Free body diagram of C Free body diagram of C

Figure 17

Balanced force system

The external forces are (i) weight of c = ? and (ii) tension in BC, T 2 T 2 .

∑ F y = T 2 - W C = 0 ⇒ W C = T 2 = 15 N ∑ F y = T 2 - W C = 0 ⇒ W C = T 2 = 15 N