How to Calculate Tension in Physics

Determine the tension at the ends of the rope. The tension in the rope is the result of its two ends being pulled. Repeat the formula “force = mass × acceleration . Assuming the rope is stretched very taut, any change in the weight or acceleration of the object causes the tension in the string to change. Don’t forget about the acceleration caused by force – even if the system is at rest, everything in the system will still be subject to this force. We have the string tension formula T = (m × g) + (m × a), where “g” is the acceleration due to gravity of the objects in the system and “a” is the specific acceleration of the body.

• In physics, to solve problems, we often hypothesize that the string is in “ideal conditions” – that is, the rope in use is very strong, has no mass or negligible mass, and cannot flex or break.
• As an example, consider a system consisting of a weight hanging on a string as shown in the figure. Both objects do not move because they are at rest. Position, we know that with the weight in equilibrium, the tension on the string must be equal to gravity. In other words, Force (F _t ) = Gravity (F _g ) = m × g.
  • Assuming a mass of 10 k, the string tension is 10 kg × 9.8 m/s ² = 98 Newtons.

Now we add the acceleration. While force is not the only factor affecting the tension in the string, any other force related to the acceleration of the object the rope is holding is equally likely. For example, if we apply a force that changes the motion of a suspended object, the force on the object’s acceleration (mass × acceleration) will be added to the value of the tension in the string.

• In our example: Let a 10 kg weight hang on a rope, but instead of before the rope was fixed to the wooden beam, now we pull the rope vertically with an acceleration of 1 m/s ² . In this case, we have to include the acceleration of the mass as well as gravity. The calculation is as follows:
  • F _t = F _g + m × a
  • F _t = 98 + 10 kg × 1 m/s ²
  • _Ft = 108 Newtons.

Add rotational acceleration. A spinning object rotating a fixed center through a string (e.g. a pendulum) will produce tension based on the centripetal force. The centripetal force also acts in addition to the tension in the string because it also “pulls” the object inward, but here instead of pulling in a straight direction, it pulls in an arc. The faster the object rotates, the greater the centripetal force. The centripetal force (F _c ) is calculated using the formula m × v ² /r where “m” is the mass, “v” is the velocity and “r” is the radius of the circle containing the arc of motion of the body.

• Since the direction and magnitude of the centripetal force changes as the object moves, so does the total tension in the string, because this force pulls the object in a direction parallel to the rope and toward the center. Also, remember that gravity is always acting in the right direction. In short, if an object is swinging in a straight line, the tension in the string will be maximum at the lowest point of the arc of motion (with a pendulum, we call it the equilibrium position), when we know that The object will move the fastest there and the fastest at the two edges.
• Still taking the example of a weight and a string, but instead of pulling, we let the weight swing like a pendulum. Assume the rope is 1.5 meters long and the weight is moving with a velocity of 2m/s when it is in equilibrium. To calculate the tension in the string in this case, we need to calculate the tension in the string due to gravity as if it were not moving at 98 Newtons, then calculate the additional centripetal force as follows:
  • F _c = m × v ² /r
  • F _c = 10 × 2 ² /1.5
  • _Fc = 10 × 2.67 = 26.7 Newtons.
  • So the total tension in the string is 98 + 26.7 = 124.7 Newtons.

You should understand that the string tension will be different at different positions of the object on the arc of motion. As noted above, both the direction and magnitude of the centripetal force of an object will change as the object moves. However, even though gravity remains constant, the string tension created by gravity will still change as usual! When the object is in an equilibrium position, gravity will act vertically and so will the tension in the string, but when the object is in a different position, these two forces will create a certain angle with each other. So the tension in the string “neutralizes” part of the gravity instead of integrating it all.

• Dividing gravity into two vectors will give you a better idea of this definition. At any point on the vertical arc of motion, the string makes an angle “θ” with the path from the center to the equilibrium position of the body. When moving, the gravitational force (m × g) will be divided into 2 vectors – mgsin(θ) asymptotically with the motion arc towards the equilibrium position. And mgcos(θ) is parallel to the tension in the string in the opposite direction. Thereby we see that the tension in the string only has to oppose mgcos(θ) – its reaction – and not the whole force of gravity (Except when the object is in equilibrium, those forces are in the same direction and direction).
• Now let the swing make an angle of 15 degrees with the vertical, moving with a speed of 1.5m/s. Then we calculate the tension as follows:
  • The tension in the string due to gravity (T _g ) = 98cos(15) = 98(0.96) = 94.08 Newton
  • Radial force (F _c ) = 10 × 1.5 ² /1.5 = 10 × 1.5 = 15 Newton
  • Total force = T _g + F _c = 94.08 + 15 = 109.08 Newtons.

Calculate additional friction force. Any object that is pulled produces a “drag” force by friction against the surface of another object (or liquid) and this force changes the tension in the string somewhat. The frictional force of the two objects in this case will also be calculated the way we usually do: The force that friction (usually denoted as F _r ) = (mu)N, where mu is the coefficient of friction where N is force exerted by two objects, or the compressive force of one object on the other. Note that static friction is different from dynamic friction – static friction is the result of bringing an object from rest to motion while dynamic friction is produced by keeping an object in its motion.

• Suppose we have a 10 kg ball but now it is dragged across the floor horizontally. Let the coefficient of dynamic friction of the floor be 0.5 and the weight initially has a constant velocity but now we add an acceleration of 1 m/s ² . This new problem has two important changes – First, we no longer calculate the tension due to gravity, because now the string tension and gravity do not cancel each other. Second, we have to add friction and acceleration. The calculation will be as follows:
  • Normal force (N) = 10 kg × 9.8 (acceleration of gravity) = 98 N
  • Dynamic Friction Force (F _r ) = 0.5 × 98 N = 49 Newton
  • Acceleration force (F _a ) = 10 kg × 1 m/s ² = 10 Newton
  • Total tension in the string = F _r + F _a = 49 + 10 = 59 Newtons.