The first image pertains to one, the second to two, and the third to 6. I\'m not
ID: 1531746 • Letter: T
Question
The first image pertains to one, the second to two, and the third to 6. I'm not entirely certain what the figures could help you with but so be it.
A box is suspended by two ropes as shown in the figure. The scales read FL = 425N at 45 degrees and FR = 601N at 60 degrees.
1. Determine the weight of the box, in N.
821 N
2. Let us first concentrate on the central value. Take F0 = 325N and E0 = 57degrees. Find the central value of Fx, in N.
177 N
Here is where I get stuck on this series of problems
3. Let ?F = 15N and ?? = 1 degree. Find Fx,min, the smallest possible value of Fx, in N.
? The answer is not 180.173N or any variant of that number
4. Find Fx,max, the largest possible value of Fx, in N.
? The answer is not 325N or any variant of that number
5. Your answers to questions 3 and 4 define a range for Fx. As seen in the lecture, the range can then be transformed into the usual error notation: central value ± error. The central value was already computed in question 2. What is the error?
? The answer is not .002 or any variant of that number.
I did figure out six.
6. What is the central value of the x-component of the net force on this system, in N?
F1 = 325 N at 57 degrees. F2=421N at 44 degrees.
325cos57+421cos(180-44) = 125.834N
7. For both forces and for both angles, the uncertainty is ?F = 15N and ?? = 1 degree. Find the corresponding error in the x-component of the net force on this system, in N. Hints: ? Propagate the uncertainty for the x-component of each force first. ? Then, propagate through the subtraction (see example in the lecture) ? Keep at least 3 decimals in all the intermediate steps
? The answer is not 14.998 or any variant of that number
45° 60°Explanation / Answer
(1) using newtons second law along y-axis
425sin45+601 sin60- w=0
w= 425sin45+601 sin60
w=821 N
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