Video: Determining Tensions in a System of Suspended Objects

Consider the baby weighed in the shown figure. What is the mass of the infant and basket if a scale reading of 55 N is observed? What is tension 𝐓₁ in the cord attaching the baby to the scale? What is tension 𝐓₂ in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?

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Video Transcript

Consider the baby weighed in the figure shown. What is the mass of the infant and basket if a scale reading of 55 newtons is observed? What is tension 𝐓 one in the cord attaching the baby to the scale? What is tension 𝐓 two in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kilograms?

In this problem, we’ll assume that the acceleration due to gravity 𝑔 is exactly 9.8 metres per second squared and we’ll also assume that the cords in the diagram are massless. We’re told in the statement that the scale records a value of 55 newtons; we’ll call that π‘Š.

In the first part of the problem, we’re asked to solve for the mass of the infant in basket; we’ll call that π‘š. We’re then asked to solve for 𝐓 one, the tension in the cord attaching the baby to the scale. Finally, we want to solve for tension 𝐓 two in the cord that attaches the scale to the ceiling, where the scale we’re told has a mass of 0.500 kilograms, which we’ll call π‘š sub 𝑠.

Let’s begin solving for part one by recalling the relationship between weight, mass, and acceleration due to gravity. Recall that an object’s weight π‘Š is equal to its mass times the acceleration due to gravity. In our instance, we’ve been told the weight of 55 newtons and want to solve for the mass of the baby in basket together.

Dividing both sides of the equation by 𝑔, that term cancels out on the right, giving us an equation that says that mass is equal to an object’s weight divided by the acceleration due to gravity. When we plug in 55 newtons for weight and 9.8 metres per second squared for 𝑔, when we perform this division, we find a mass to two significant figures of 5.6 kilograms; that’s the mass of the baby and basket.

Now, we turn to solving for 𝐓 one, the tension in the cord between the baby and the scale. If we model the baby in basket as a point and draw the forces acting on that point, those forces are the weight force π‘Š and 𝐓 one, the tension force that acts up. Because the baby is in equilibrium, these forces are equal in magnitude. So 𝐓 one the tension force in the cord between the baby and the scale is equal to 55 newtons. That is the tension in the lower of the two chords.

Finally, we want to solve for 𝐓 two the tension in the cord that attaches the scale to the ceiling. If we model the scale as a point and draw the forces acting on it, we have 𝐓 two acting up, 𝐓 one acting down, and the weight force of the scale acting down as well. Because the scale is in equilibrium, we can write that the magnitude of 𝐓 two is equal to 𝐓 one plus π‘Š sub 𝑠, the weight force of the scale.

The weight force of the scale is equal to the mass of the scale times 𝑔. We’ve been given the mass of the scale, solved for 𝐓 sub one and no 𝑔. So we can now plug in for each of these three values to solve for 𝐓 two. 𝐓 two is equal to 55 newtons plus 0.500 kilograms times 9.8 metres per second squared. To two significant figures, this equals 60 newtons. That is the tension in the cord attaching the scale to the ceiling.

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