Hey, I need help working these problems. Thanks you.

1) A 50-m tape is used to measure between two points. The average weight of the tape per meter is 0.320 N. If the measured distance is 48.888 m, with the tape supported at the ends only and with a tension of 100 N, find the corrected distance.

2) A100-ft steel tape weighing 1.8 lb and supported only at the ends with a tension of 24 lb is used to measure a distance of 471.16 ft. What is the distance corrected for sag?

3) A distance of 72.55 ft is recorded using a steel tape supported only at the ends with a tension of 15 lb and weighing 0.016 lb per foot. Find the distance corrected for sag.

The correction for sag, Cs, unsupported between two given points L apart, and subject to a pull (tension) P, is given by

Cs=ω²L³/(24P²)
where ω is the weight of the tape/unit length.

In the given case,
L=48.88 m
ω=0.32N/m
P=100N

Cs=0.32²*48.88³/(24*100²)
=0.0498m
Check:
percentage correction
=0.0498/48.88=0.1%
This is a rather substantial correction, but in consideration of the span of 49m, it is understandable.

The two other problems are similar, using different units. The percentages correction are about 1/4 of that of problem (1).

Notes:
(1) It is important that the tension applied is the specified tension for the tape, or else an additional correction for tension has to be applied.
(2) This kind of measurement is becoming less and less day-to-day work, as the new instruments in total stations can measure to 2ppm±2mm in a matter of a second, up to 500 m or more.

Sure, I can help you with these problems. Let's tackle them step by step:

1) To find the corrected distance, we need to account for the sag in the tape. The sag in the tape can be calculated using the formula:

Sag = (T^2 / (8 * w)) * (1 + (L / T)^2)

Where:
T = tension (100 N)
w = weight per meter (0.320 N/m)
L = measured distance (48.888 m)

First, let's calculate the sag:
Sag = (100^2 / (8 * 0.320)) * (1 + (48.888 / 100)^2)
Sag ≈ 190.617 m

Next, we need to subtract the sag from the measured distance to find the corrected distance:
Corrected distance = Measured distance - Sag
Corrected distance = 48.888 m - 190.617 m
Corrected distance ≈ -141.729 m

Note that the negative sign indicates that the corrected distance is shorter than the measured distance.

2) The sag in the tape for this problem can be calculated using the same formula as in problem 1:

Sag = (T^2 / (8 * w)) * (1 + (L / T)^2)

Where:
T = tension (24 lb)
w = weight per foot (1.8 lb/100 ft)
L = measured distance (471.16 ft)

First, let's convert the given measurements:
T = 24 lb
w = 1.8 lb / 100 ft = 0.018 lb / ft
L = 471.16 ft

Next, let's calculate the sag:
Sag = (24^2 / (8 * 0.018)) * (1 + (471.16 / 24)^2)
Sag ≈ 221.537 ft

Finally, subtract the sag from the measured distance to get the corrected distance:
Corrected distance = Measured distance - Sag
Corrected distance = 471.16 ft - 221.537 ft
Corrected distance ≈ 249.623 ft

3) The sag in the tape can be calculated using the same formula as in the previous problems:

Sag = (T^2 / (8 * w)) * (1 + (L / T)^2)

Where:
T = tension (15 lb)
w = weight per foot (0.016 lb/ft)
L = measured distance (72.55 ft)

First, let's calculate the sag:
Sag = (15^2 / (8 * 0.016)) * (1 + (72.55 / 15)^2)
Sag ≈ 183.649 ft

Finally, subtract the sag from the measured distance to find the corrected distance:
Corrected distance = Measured distance - Sag
Corrected distance = 72.55 ft - 183.649 ft
Corrected distance ≈ -111.099 ft

Similarly to problem 1, the negative sign indicates that the corrected distance is shorter than the measured distance.

To solve these problems, we need to account for the sag or deformation in the tape due to its own weight. We can use the formula for the correction due to sag:

Correction = (Weight per unit length × Length × Length) / (8 × Tension)

where Weight per unit length is the weight of the tape per meter or foot, Length is the measured distance, and Tension is the tension applied to the tape.

Let's solve each problem step by step:

1) Find the corrected distance when the measured distance is 48.888 m. The weight per meter of the tape is 0.320 N, and the tension is 100 N.

Correction = (0.320 N/m * 48.888 m * 48.888 m) / (8 * 100 N)

2) Find the corrected distance when the measured distance is 471.16 ft. The weight per foot of the steel tape is 1.8 lb, and the tension is 24 lb.

Correction = (1.8 lb/ft * 471.16 ft * 471.16 ft) / (8 * 24 lb)

3) Find the corrected distance when the measured distance is 72.55 ft. The weight per foot of the steel tape is 0.016 lb, and the tension is 15 lb.

Correction = (0.016 lb/ft * 72.55 ft * 72.55 ft) / (8 * 15 lb)

By calculating these corrections, you can find the corrected distances in each problem.