For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can’t be stretched or broken. As an example, let’s consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g. Assuming a 10 kg weight, then, the tension force is 10 kg × 9. 8 m/s2 = 98 Newtons.
For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can’t be stretched or broken. As an example, let’s consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g. Assuming a 10 kg weight, then, the tension force is 10 kg × 9. 8 m/s2 = 98 Newtons.
Let’s say that, in our example of the 10 kg weight suspended by a rope, that, instead of being fixed to a wooden beam, the rope is actually being used to pull the weight upwards at an acceleration of 1 m/s2. In this case, we must account for the acceleration on the weight as well as the force of gravity by solving as follows: Ft = Fg + m × a Ft = 98 + 10 kg × 1 m/s2 Ft = 108 Newtons.
Since the direction and magnitude of centripetal force changes as the object on the rope moves and changes speeds, so does the total tension in the rope, which always pulls parallel to the rope towards the central point. Remember also that the force of gravity is constantly acting on the object in a downward direction. So, if an object is being spun or swung vertically, total tension is greatest at the bottom of the arc (for a pendulum, this is called the equilibrium point) when the object is moving fastest and least at the top of the arc when it is moving slowest. [4] X Research source Let’s say in our example problem that our object is no longer accelerating upwards but instead is swinging like a pendulum. We’ll say that our rope is 1. 5 meters (4. 9 ft) long and that our weight is moving at 2 m/s when it passes through the bottom of its swing. If we want to calculate tension at the bottom of the arc when it’s highest, we would first recognize that the tension due to gravity at this point is the same as when the weight was held motionless - 98 Newtons. To find the additional centripetal force, we would solve as follows: Fc = m × v2/r Fc = 10 × 22/1. 5 Fc =10 × 2. 67 = 26. 7 Newtons. So, our the total tension would be 98 + 26. 7 = 124. 7 Newtons.
Breaking gravitational force up into two vectors can help you visualize this concept. At any given point in the arc of a vertically swinging object, the rope forms an angle “θ” with the line through the equilibrium point and the central point of rotation. As the pendulum swings, gravitational force (m × g) can be broken up into two vectors - mgsin(θ) acting tangent to the arc in the direction of the equilibrium point and mgcos(θ) acting parallel to the tension force in the opposite direction. Tension only has to counter mgcos(θ) - the force pulling against it - not the entire gravitational force (except at the equilibrium point, when these are equal). Let’s say that when our pendulum forms an angle of 15 degrees with the vertical, it’s moving 1. 5 m/s. We would find tension by solving as follows: Tension due to gravity (Tg) = 98cos(15) = 98(0. 96) = 94. 08 Newtons Centripetal force (Fc) = 10 × 1. 52/1. 5 = 10 × 1. 5 = 15 Newtons Total tension = Tg + Fc = 94. 08 + 15 = 109. 08 Newtons.
Let’s say that our 10 kg weight is no longer being swung but is now being dragged horizontally along the ground by our rope. Let’s say that the ground has a kinetic friction coefficient of 0. 5 and that our weight is moving at a constant velocity but that we want to accelerate it at 1 m/s2. This new problem presents two important changes - first, we no longer have to calculate tension due to gravity because our rope isn’t supporting the weight against its force. Second, we have to account for tension caused by friction, as well as that caused by accelerating the weight’s mass. We would solve as follows: Normal force (N) = 10 kg × 9. 8 (acceleration from gravity) = 98 N Force from kinetic friction (Fr) = 0. 5 × 98 N = 49 Newtons Force from acceleration (Fa) = 10 kg × 1 m/s2 = 10 Newtons Total tension = Fr + Fa = 49 + 10 = 59 Newtons.
Note that, usually, physics problems assume ideal pulleys - massless, frictionless pulleys that can’t break, deform, or become separated from the ceiling, rope, etc. that supports them. Let’s say we have two weights hanging vertically from a pulley in parallel strands. Weight 1 has a mass of 10 kg, while weight 2 has a mass of 5 kg. In this case, we would find tension as follows: T = 2g(m1)(m2)/(m2+m1) T = 2(9. 8)(10)(5)/(5 + 10) T = 19. 6(50)/(15) T = 980/15 T = 65. 33 Newtons. Note that, because one weight is heavier than the other, all other things being equal, this system will begin to accelerate, with the 10 kg moving downward and the 5 kg weight moving upward.
Let’s say we have a system with a 10 kg weight (m1) hanging vertically connected by a pulley to a 5 kg weight (m2) on a 60 degree ramp (assume the ramp is frictionless). To find the tension in the rope, it’s easiest to find equations for the forces accelerating the weights first. Proceed as follows: The hanging weight is heavier and we’re not dealing with friction, so we know it will accelerate downward. The tension in the rope is pulling up on it, though, so it’s accelerating due to the net force F = m1(g) - T, or 10(9. 8) - T = 98 - T. We know the weight on the ramp will accelerate up the ramp. Since the ramp is frictionless, we know that the tension is pulling it up the ramp and only its own weight is pulling it down. The component of the force pulling it down the ramp is given by sin(θ), so, in our case, we can say that it’s accelerating up the ramp due to the net force F = T - m2(g)sin(60) = T - 5(9. 8)(. 87) = T - 42. 63. [7] X Research source Acceleration of the two weights are the same, thus we have (98 - T)/m1 = (T - 42. 63) /m2. After a little trivial work to solve this equation, finally we have T = 60. 96 Newton.
Let’s say in our Y-shaped system that the bottom weight has a mass of 10 kg and that the two upper ropes meet the ceiling at 30 degrees and 60 degrees respectively. If we want to find the tension in each of the upper ropes, we’ll need to consider each tension’s vertical and horizontal components. Nonetheless, in this example, the two ropes happens to be perpendicular to each other, making it easy for us to calculate according to the definitions of trigonometric functions as follows: The ratio between T1 or T2 and T = m(g) is equal to the sine of the angle between each supporting rope and the ceiling. For T1, sin(30) = 0. 5, while for T2, sin(60) = 0. 87 Multiply the tension in the lower rope (T = mg) by the sine of each angle to find T1 and T2. T1 = . 5 × m(g) = . 5 × 10(9. 8) = 49 Newtons. T2 = . 87 × m(g) = . 87 × 10(9. 8) = 85. 26 Newtons.
