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no idea how to approach this qs - physics - please help?
This problem concerns the concept of tension in a rope. Consider a rope subjected to a pulling force on its two ends as shown. (Figure 1) The rope is stationary. An arbitrary point P divides the rope into a left-hand segment L and a right-hand segment R.
1.Assume that segment R exerts a force of magnitude T on segment L. What is the magnitude FRL of the force exerted on segment R by segment L?
Give your answer in terms of T and other constants such as g.
2.Now imagine two points, Q and P, that divide the rope into segments L, M ,and R. (Figure 2) The rope remains stationary. Assume that segment L exerts a force of magnitude FLM on segment M. What is the magnitude FRM of the force exerted by segment R on segment M?
Give your answer in terms of FLM and constants such as g.
3.Now consider a rope that, unlike those usually studied in mechanics problems, actually has a significant mass m. The tension at the right end of this rope is T2 and that at the left end is T1. (Figure 3) The rope has an acceleration arope to the right.
Complete the following equation for the force on the section of the rope of mass m, taking the positive direction to be to the right.
Give your answer in terms of T1, T2, and constants such as g.
4.Which of the following phrases, if they appear in a problem, allow you to assume that T2=T1 in a horizontally oriented rope?
The rope is massless.
The rope is moving at constant speed.
The rope is stretched with negligible sag.
kind regards,
2 Answers
- 7 years ago
1. The question says FLR = T, since FLR and FRL are action/reaction pairs (Newton's 3rd Law), FRL will also equal T
So, the answer is: T
2. If you draw a force diagram of section M, you should be able to tell that both FLM = FRM
So, the answer is: FLM
3. Use Newton's Second Law here. Since the right is the positive direction, (and the force of T2 is going right), T2 will be positive. Following this line of thinking, since T1 is going left, it will be negative...
So, the answer is: T2 - T1
4. The reason why a massless rope would cause T1=T2 can be seen using Newton's 2nd law: F=ma. The force F is given by T2−T1. Since m=0 then ma=0 for any value of a. This means that the force must be zero (F=ma=0). Therefore, T1=T2.
And when a rope is moving at constant speed, the net force on the rope will be 0, therefore T1=T2
So, the first and second statements allow us to assume T1=T2
Source(s): I'm doing my physics homework now as well ;)
