Weighing with A 400 N Scale and 80 N Spring Dynamometer

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SUMMARY

The discussion centers on a mechanics problem involving a 400 N scale and an 80 N spring dynamometer to determine a student's mass. When the dynamometer registers 76 N, the scale reads 268 N, leading to the equation for mass: m = (T1 + T2 + NA) / g. The tension T2 is established as four times T1, confirming T2 = 4T1. This setup allows for a range of weights to be measured, depending on the equilibrium conditions of the system.

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Homework Statement



A former student of mechanics wishes to weigh himself but has access only to a scale A with capacity limited to 400 N and a small 80 N spring dynamometer B. With the rig shown he discovers that when he exerts a pull on the rope so that B registers 76 N, the scale A reads 268 N. What is his mass ##m##? Also, what range of weight could be measured by this system?

Homework Equations



Since the student is in equilibrium:

$$\sum F_y = 0$$

The Attempt at a Solution



The normal reaction acting on the student is 268 N. There are two tensions acting on the student; ##T_1## and ##T_2## (see diagram).
##T_1## is clearly 76 N. However, I can't seem to find ##T_2##. How do I assign a value to ##T_2## based on the arrangement of the ropes shown in the diagram?
Once found:

$$m = \frac{T_1 + T_2 + N_A}{g}$$

Update: I tried finding ##T_2## by first noting that the tension throughout the string from B to the ceiling is 76 N. Applying the equilibrium condition for the two pulleys with strings directly attached to the triangular support to which the rope in question is attached. We now have ##T_2 = 4T_1##.
Is this correct?
 

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MohammedRady97 said:

Homework Statement



A former student of mechanics wishes to weigh himself but has access only to a scale A with capacity limited to 400 N and a small 80 N spring dynamometer B. With the rig shown he discovers that when he exerts a pull on the rope so that B registers 76 N, the scale A reads 268 N. What is his mass ##m##? Also, what range of weight could be measured by this system?

Homework Equations



Since the student is in equilibrium:

$$\sum F_y = 0$$

The Attempt at a Solution



The normal reaction acting on the student is 268 N. There are two tensions acting on the student; ##T_1## and ##T_2## (see diagram).
##T_1## is clearly 76 N. However, I can't seem to find ##T_2##. How do I assign a value to ##T_2## based on the arrangement of the ropes shown in the diagram?
Once found:

$$m = \frac{T_1 + T_2 + N_A}{g}$$

Update: I tried finding ##T_2## by first noting that the tension throughout the string from B to the ceiling is 76 N. Applying the equilibrium condition for the two pulleys with strings directly attached to the triangular support to which the rope in question is attached. We now have ##T_2 = 4T_1##.
Is this correct?
Yes, T2 = 4T1
 

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