Tension formula-Rope pulling blocks horizontally with kinetic friction involved

When we pull a block using a rope there is tension induced in the rope in the opposite direction. For ease of calculation, we generally assume that the rope is massless and the surface is frictionless. In our previous blog post, we had calculated the formula for tension for a similar case but without frictional force. But, in an ideal case, there will always be a small force of kinetic friction that opposes the motion of the blocks. If the force applied on the block is less than the limiting friction, then static friction will act on the blocks, and the blocks will be at rest.

The formula for tension in the case of a single block

A block of mass m1 is pulled with acceleration a. There is a tension T developed in the rope. There will be kinetic friction between the surface and the block since it is in motion.

The kinetic friction will be equal to μ_k * m1 * g

So the equation for tension will be,

Fnet = tension – friction

m1*a = T – μ_k * m1 * g

T = m1*a + μ_k * m1 * g

T = m1(a + μ_kg)

This formula for tension is consistent with general observation, as the tension has to be equal to the applied force which here is the force applied plus the frictional force.

The formula for tension in the case of dual blocks connected by a rope

In the problem below we have a force pulling two blocks attached by a piece of rope. Now, we have to find the formula for tension in the rope. It is also given that the coefficient of friction between the surface and the block is μ_k.

Before you proceed you can consider reading the following blog posts to get a deeper understanding of tension and friction:

• What is the frictional force?
• What is kinetic friction?
• What is tension?
• Formula for tension

First, let us calculate the value of friction acting on the system:

Friction(f_k) =  μ_k N = μ_k*(m_totalg)
f_k = μ_k(m₁+m₂)g

Now, this is friction is the friction of the entire system(considering both blocks together) for a force F with mass M1+M2.

Now, let us calculate the acceleration of the system:

Fnet = F – friction 

acceleration(a) = F/Total mass

a = [F-μ_k(m₁+m₂)g]/(m₁+m₂)

Now, since we have calculated the formula for the acceleration of the system, we can look at the free-body diagram of both blocks.

Tension equation for block1

The forces acting on block1 are tension and frictional force. The summation of these forces should be equal to the net force.

Fnet = T – friction

T= Fnet + friction

T = μ_k*m₁*g + m₁*a ——–(1)

Tension equation for block2

The forces acting on block 2 are applied force F, frictional force, and tension force. The summation of these forces should be equal to the net force.

Fnet = F – T – friction

T = F- friction – Fnet

T = F – μ_k*m₂*g – m₂*a ———-(2)

Formula for tension

In case of single block:

T = m1(a + μ_kg)

For case of dual blocks:

Using free-body diagram of block1:

T = μ_k*m₁*g + m₁*a

Using free-body diagram of block2:

T = F – μ_k*m₂*g – m₂*a  

Using the above two equations we can find the values for tension in a similar case.

For cases where there are three or blocks, follow the same steps:

• Calculate the acceleration of the system first.
• Calculate the friction of the system
• Use a free body diagram of each block to solvent for tension in each rope.

Numerical problem on the calculation of tension force in a rope pulling blocks with friction

1)A block of mass 1 Kg is pulled horizontally with a force of 40 N. The coefficient of friction is given as 0.25, Find the tension in the rope.

First, let us find the acceleration of the system:

F = m*a

40 = 1* a

Acceleration(a) = 40 m/s2

Let us use the formula for the tension we derived.

T = m1(a + μ_kg)

T = 1(40 + 0.25*9.8)

T = 42.45 N

2) Two blocks with masses of 3 Kg and 2Kg are pulled horizontally across a surface with a coefficient of kinetic friction of 1. Find the tension in the wire connecting block1 and block2.

Here in this problem, the coefficient of friction is equal to 1. This is very high friction, that is why you see a very low acceleration. We will apply the formula for tension as above and calculate the value of tension in the rope. We will verify using both blocks.

First, the friction of the system is calculated which is equal to 49 N. Since this value is less than the force of 50N, the system will move. If the friction value is greater than the applied force then the system will not move. It will be a case of static friction.

Now the net acceleration of the system is calculated taking into consideration the frictional force acting on the system. The acceleration comes out to be 0.2m/s2.

Apply the formula for tension for block1:

Fnet = T – friction

0.4 = T – 19.6

T = 20 N

Apply the formula for tension for block2:

Fnet = 50 – T – friction

0.6 = 50 – T – 29.4

T = 20 N

From both the formulas we find that the value of tension force is the same. Hence our tension formula is verified to be correct.

Some real-life examples:

Chariots are good examples of ropes pulling objects horizontally. Though there are no blocks here. Chariots use wheels to reduce friction. The sliding friction is converted into rolling friction to reduce the pulling force required.

Sled dogs are another example of ropes using to pull objects. It is a common mode of transportation in ice-cold regions. The tension in the rope enables the dogs to pull weights. The friction involved here can be slightly less as compared to normal dry ground.

See also:

• Tension formula- Tug of war
• Tension Formula-Tension in a rope pulling blocks horizontally
• Formula For Tension
• Tension formula-Rope 
• Tension formula: Tension in a vertically suspended wire with a weight
• Tension elevator
• Tension formula circular motion
• Pulley system