We discuss problems, which highlight certain aspects of the study leading to the balanced force system. The questions are categorized in terms of the characterizing features of the subject matter :

Problem 1 : 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 strings.

Figure 1

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 2

Balanced force system

A preliminary assessment of forces suggests that analysis of forces on block will provide value for the unknown, T 1 T 1 . Hence, we first analyze force on the block.

Free body diagram of block Free body diagram of block

T 1 = 100 N T 1 = 100 N

Figure 3

Free body diagram

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 " T 1 T 1 " acts up with respect to the body, whereas " T 1 T 1 " 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 : Find the force, F, required to keep the block stationary on an incline of angle "θ" having friction-less surface as shown in the figure.

Figure 4

Balanced force system

Solution : We can either have (a) axes in horizontal and vertical directions or (b) parallel to incline and perpendicular to it. Which of the two is better suited here ? In this case, one force (mg) is along vertical direction, whereas other external force (F) is along horizontal direction. As such, it is advantageous to have a horizontal and vertical axes as two of three forces are along the coordinate axes.

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

The external forces on the block are (i) Force, F (ii) Weight, mg and (iii) Normal force, N.

Figure 5

Free body diagram

∑ 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 - mg = 0 ⇒ mg = N cos θ ∑ F y = N cos θ - 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 4 : 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 6

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 and mass-less, whereas pulley is not mass-less. However, pulley is static. As such, there is no torque involved. Hence, tension in the string all through out is same. From the consideration of block, we see that tension in the string is equal to the weight of the block i.e. Mg.

Now, the external forces on pulley are (i) Horizontal tension "T" (ii) Weight, mg, of the pulley (iii) Vertical Tension, T, and (iv) force, F applied by clamp “B”.

Figure 7

Free body diagram

∑ F x = F x - T = F x - Mg = 0 ⇒ F x = Mg ∑ F x = F x - T = 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 = g { M 2 + ( M + m ) 2 } F = ( F x 2 + F y 2 ) = { ( Mg ) 2 + ( M + m ) 2 g 2 } ⇒ F = g { M 2 + ( M + m ) 2 }

Problem 5 : Two blocks "A" and "B", weighing 20 N and 10 N respectively are in contact with each other. If the blocks are at rest, then find the force "F" and the normal reactions between all contact surfaces.

Figure 8

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) external force of 20 N

Figure 9

Free body diagram

∑ 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 10

Free body diagram

∑ F x = F cos 30 0 - N 3 = F cos 30 0 - N 1 = F cos 30 0 - 20 = 0 ⇒ 20 = F cos 30 0 ⇒ F = 20 cos 30 0 = 20 √ 3 2 = 40 √ 3 N ∑ F x = F cos 30 0 - N 3 = F cos 30 0 - N 1 = F cos 30 0 - 20 = 0 ⇒ 20 = F cos 30 0 ⇒ F = 20 cos 30 0 = 20 √ 3 2 = 40 √ 3 N

and

∑ F y = N 4 - F sin 30 0 - 20 = 0 ⇒ N 4 = F sin 30 0 + 20 = 40 √ 3 X 1 2 + 20 = 20 √ 3 + 20 = 20 ( 1 + √ 3 √ 3 ) N ∑ F y = N 4 - F sin 30 0 - 20 = 0 ⇒ N 4 = F sin 30 0 + 20 = 40 √ 3 X 1 2 + 20 = 20 √ 3 + 20 = 20 ( 1 + √ 3 √ 3 ) N

Problem 6 : 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. If the block system is in equilibrium, find the weight of C and tensions in the two strings.

Figure 11

Balanced force system

Solution : There are two strings. Hence, the tensions in the strings will 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 C. 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 12

Free body diagram

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 not analyze forces in y – direction as we are not required to determine normal force N 1 N 1 and it is not expected to be 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 13

Free body diagram

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

Author wishes to thank Scott Kravitz, Programming Assistant, Connexions for making suggestion to remove error in the module.