Tension

Introduction
toc Tension is a force exerted by a string. Since it is a force, it can be drawn on a free body diagram. The force of tension always points parallel to the string; a string cannot push, only pull.

There is no equation for finding the tension of a string. It is important to know that tension does not depend on the string's characteristics. So the length, material, or coarseness plays no part in finding tension. You need to exert the same force to pull the same mass, whether you're using a long, thin string, or a short, thick piece of yarn. Tension has to be found using the other clues in the problem.

Tension is usually found in combination with other forces in Newton's Second Law, ΣF = ma. This is simply because there is no exclusive equation for finding tension. Drawing free body diagrams is especially important in these problems, since other acting forces are usually essential pieces of information to finding the tension.

**Problem #1**
A 5 kg block is sitting on a horizontal, frictionless surface. A string is attached to it, and the block with an acceleration of 2 m/s 2 to the right. What is the tension in the string?

**Problem #2**
A 15 kg block is sitting on a frictionless incline ramp with a base angle of 30°. A string is attached to it, and a person is holding the block in equilibrium. What is the tension in the string?

**Problem #3**
A boy swings a 3 kg ball around in a circle above his head and parallel to the ground. If the string is 2 m, and the ball travels the circular path at a tangential velocity, //v//, of 10 m/s, then what is the tension in the string?

**Problem #4**
Two weights, masses m and M, are attached by the same string and the string is threaded around a frictionless pulley. What is the acceleration of the masses after the weights are allowed to fall?

**Problem #5**
Two blocks, masses m and M, are sitting on a frictionless, horizontal surface, and are attached by a string. A force F is exerted on mass M, and the system accelerates to the right. If the force remains in the same direction, but is increased to 2F, what happens to the tension of the string connecting the blocks in the transformation of the force from F to 2F? A) the tension stays the same B) the tension increases C) the tension decreases

Challenge Problem
A 5 kg block is sitting on an incline ramp with base angle of 30°. If the coefficient of kinetic friction between the block and the ramp is 0.25, and a string pulls the block uphill with an acceleration of 2 m/s 2, what is the tension in the string?

Problem #1
Because the acceleration is to the right, the net force has to be pointing to the right. The normal force and force of gravity cancel out, so the net force on the block is the tension, to the right. Σ F = T = ma T = (5 kg)(2 m/ s 2 ) T = 10 N

Problem #2
The forces acting on the block are as drawn in the below figure. The downward force of gravity weight is split into two components: the force that guides the block downhill, and the force keeping the block on the ramp. The force keeping the block on the ramp is counterbalanced by the normal force, so the net force on the block without the string is the downhill force. The downhill force can be found with the expression mg sin θ. For the block to be in equilibrium, the tension has to be equal to the downhill force in magnitude, but point in the opposite direction. T = mg sin θ T = (15 kg)(10 m/ s 2 )(sin 30°) T = 150 N (1/2)
 * T = 75 N**

Problem #3
This is a problem dealing with centripetal force, because it keeps the ball traveling in a circle. The net force on the ball is the tension in the string, so the tension is equal to the centripetal force. F c = (mv 2 )/r = T Although the tension in this case depends on the length of the string, r, tension in general does not rely on the string's characteristics. For instance, swinging this 3 kg ball with a 2 m piece of yarn would not alter the tension. Exceptions are only seen with centripetal force problems and the length of the string. T = [(3 kg) (10 m/s) 2 ] / 2 m
 * T = 150 N**

Problem #4
The heavier weight would cause the system to fall in its direction. So if you designate direction of motion as positive, your free body diagrams should look something like this:

Because the two weights are connected by a string, the acceleration of mass m is the same as that of mass M; the string makes them move together. So setting up Newton's Second Law for both weights: Mg - T = Ma T - mg = ma The acceleration a is the same value for both equations because the weights are connected by the string. Adding the two equations yields: Mg - T + T - mg = Ma + ma Mg - mg = g(M - m) = a(M + m)
 * a = [g(M - m)] / (M + m)**

Problem #5
Force F accelerates the system to the right. With force 2F, the system would accelerate faster to the right. The net force on mass m is tension T. If mass M accelerates at **a**, mass m has to accelerate at **a** as well. If mass M accelerates at **b**, mass m has to accelerate at **b** as well. The two blocks move together because they are connected by the string. Because M moves faster with 2F than F, m has to move faster as well. So in order for m to move faster, the net force on mass m, or tension T, has to increase as well.
 * The answer is (B).**

Challenge Problem Solution
There are four forces on the block: the force of gravity downwards, the normal force from the ramp, the force of kinetic friction downhill (opposes the uphill motion), and of course, tension uphill. The force of gravity can be split up into two components, the downhill force and the force holding the block to the ramp. The normal force and the force holding the block to the ramp cancel out. The only forces acting parallel to the motion are: the force of kinetic friction downhill, the downhill force, and the tension uphill. So combining the forces in each direction, uphill and downhill, the free body diagram becomes: So using Newton's Second Law, you can set up the equation: ΣF = T - ( μ mg cos θ + mg sin θ) = ma T - μ mg cos θ - mg sin θ = ma T = ma + μ mg cos θ + mg sin θ

T = (5 kg)(2 m/s 2 ) + (0.25)(5 kg)(10 m/s 2 )(cos 30°) + (5 kg)(10 m/s 2 )(sin 30°) T = 10 N + 10.825 N + 25 N
 * T = 45.825 N**