Question

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F⃗ . (Figure 1) The magnitude of the tension in the string between blocks B and C is T = 3.00 N . Assume that each block has mass m = 0.400 kg .

What is the magnitude F of the force?

What is the tension T_AB in the string between block A and block B?

Answer 1

Concepts and reason

The concept used to solve this problem is Newton’s second law of motion.

The tension in each rope can be calculated by drawing a free body diagram of the system with all the forces acting on the system.

The whole system consists of three blocks and they are connected by two ropes and are being pulled by another rope. The free body diagram must be made separately for every block.

The tension in each rope can be calculated by balancing all the forces acting on each block during the motion.

The force acting on the block can be calculated by finding the tensions in every rope.

Fundamentals

The Newton’s first law of motion states that when a particle is at rest or in uniform motion, the net force (sum of all the forces) acting on it must be zero.

$\sum \vec F = 0$

A particle is not said to be in equilibrium when it is accelerating under the influence of an external force. The net force ( $\sum \vec F$ ) acting on the particle in this case is not equal to zero but is equal to the product of its mass (m) and the acceleration ( $\vec a$ ).

$\sum \vec F = m\vec a$

(a)

The figure 1 represents a free-body diagram of a system of three blocks which are connected by two different ropes. The system of three blocks moves with a constant acceleration $\vec a$ under the influence of force $\vec F$ . The blocks are connected by two different ropes.

Refer figure 1 and balance the forces acting on block C.

$F - {T_1} = ma$

Refer figure 1 and balance all the forces acting on block B.

${T_1} - {T_2} = ma$

Refer figure 1 and balance all the forces acting on block A.

${T_2} = ma$

Add all the above equations.

$\begin{array}{c}\\F - {T_1} + {T_1} - {T_2} + {T_2} = ma + ma + ma\\\\F = 3ma\\\end{array}$

Write the balanced equation of forces acting on block A.

${T_2} = ma$

Write the balanced equation of forces acting on block B.

${T_1} - {T_2} = ma$

Substitute ma for ${T_2}$ in the above equation.

$\begin{array}{c}\\{T_1} - ma = ma\\\\2ma = {T_1}\\\end{array}$

Substitute 3.00 N for ${T_1}$ in the above equation.

$\begin{array}{c}\\2ma = 3.00{\rm{ N}}\\\\ma = \left( {\frac{{3.00{\rm{ N}}}}{2}} \right)\\\\ = 1.50{\rm{ N}}\\\end{array}$

The tension in the rope between wire A and B is as follows:

${T_2} = ma$

Substitute 1.50 N for ma in the above equation.

${T_2} = 1.50{\rm{ N}}$

(b)

Determine the force acting on the system.

Refer figure 1 and balance the forces acting on block C.

$F - {T_1} = ma$

Substitute 3.00 N for ${T_1}$ and 1.50 N for ma in the above equation.

$\begin{array}{c}\\F - \left( {3.00{\rm{ N}}} \right) = \left( {1.50{\rm{ N}}} \right)\\\\F = \left( {1.50{\rm{ N}}} \right) + \left( {3.00{\rm{ N}}} \right)\\\\ = 4.50{\rm{ N}}\\\end{array}$

Ans: Part a

The tension in the rope between A and B is 1.50 N.

Answer 2

F= 4.5N and the

T_AB=1.5N

Answer 3

F = 3ma

Tbc = 2ma

Tab = ma

Answer 4