What input voltage will produce full-scale deflection in...

Key Concepts

Full-Scale Deflection

Full-scale deflection is the maximum reading or indication an analog instrument can display, determined by the intrinsic properties of the measurement device. It is the benchmark against which the instrument's response to input is calibrated, ensuring that when the specified limit is reached, the device indicates its full range of motion.

Input Voltage Scaling

Input voltage scaling involves establishing a relationship between the external signal applied to a circuit or instrument and the resulting response, such as achieving full-scale deflection. This concept relies on understanding how the input voltage is divided or conditioned by the circuit elements to match the instrument’s operational limits.

Voltage Divider Principle

The voltage divider principle is a fundamental concept in circuit analysis where a voltage input is split among components in proportion to their resistances. This technique is crucial in designing circuits for measurement instruments to ensure that the input voltage is appropriately reduced or shifted so that the device responds correctly, especially at full-scale conditions.

Calibration of Measuring Instruments

Calibration of measuring instruments is the process of setting or adjusting the output of an instrument to ensure its readings are accurate and align with known standards. This is essential when determining the precise input voltage that corresponds to full-scale deflection, thereby ensuring reliable and repeatable measurements.

Transcript

00:01 And this problem will be talking about forces in motion, more specifically tension forces.

00:06 So the goal in this problem is to solve for each force in each string or cord.

00:13 And i have them all labeled here in blue on this tiger.

00:16 So we're going to start with the first one labeled t because it is the only one that has a value attached to it, which is the 10 kilogram mass.

00:25 So in that cord, there's only one force on it, which is the 10 kilogram mass at the bottom.

00:36 So we can take that and solve for that force in t.

00:45 So we do 10 times, oh my goodness, times 9 .8, and we get 98 newtons for t.

01:01 Okay, the next thing would be to look at this point and look at the forces coming off of it.

01:11 So i'm going to draw that right here.

01:14 So we have down, which is t, we have this, which i labeled as t -a -1, and this as t -b -1.

01:28 Okay, so the first step is we're going to make equations for the sum of the forces in the x direction, of some of the forces in the y directions.

01:37 So, let's do some in the x direction first.

01:43 So that sum of forces in the x is going to be equal to zero because it's going to not be moving.

01:49 So it's static, so the net force is zero.

01:52 And then because these angles are 30, i'm just going to redraw the dash lines.

01:58 And what adds up to 30 is 60.

02:01 So this angle right here is 60.

02:04 And i'm going to use that because it makes a triangle that i think it's easier to see.

02:13 So we get tb1 cosine 60 minus t .a .1 cosine 60.

02:31 Okay.

02:41 We get tb1 equals t a1 cosine 60 over cosine 60, which these cancel, and we get that tb1 equals ta1, which makes sense because they're the same angle coming off from that one point with the one force on the bottom, so they're going to be equal because they have the equal angles.

03:11 So then let's write the equation for the sum of the forces in the y direction, which is also equal to 0.

03:19 And let's write it as t b is equal to t a.

03:22 So if we have, well first i'll write it with tb1 and then substitute...

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