Lifting Boom A untitled folder LON-CAPA Lit a Dominique Parry (Student-section:

Question

Lifting Boom A untitled folder LON-CAPA Lit a Dominique Parry (Student-section: 730) PHY231C, Summer 2016-Introductory Physics i g 2016- Introductory Physics I Messages Courses Help Logout )-Lifting TImer Notes Evaluate Feedback Print Info y a cable attached to the wall and the lower end by a pivot (marked X) Merwe ean Course Contents »... course contents » * Homework set #7 due 6/14 at 5PM Homework on the same wall. Calculate the tension in the cable. 12 16 0 123 4567 8 9 10 11 12 13 14 15 16 Horizontal position, x, (m) 2993.6 N one of t he tic marks.) You'll need to get the various positions from the graph. Many are exactly on Submit Answer Incorrect. Tries 7/20 Previous Tries on what is marked as NEW onger new Throaderd View Chronologlical View Other Viewsn My. NEW Anonymous 1 Reply (Sun Jun 12 10:57:16 pm 2016 (EDT)) ces t the torque around pivot and setting

Explanation / Answer

m = mass of the crate = 179.5 kg
m = mass of the boom = 89.9 kg
g = acceleration by gravity = 9.81 m/s²
L = length of the boom
= angle of the boom to horizontal
= angle of the cable to horizontal

From the image you can see that
tan() = 4/13 = 0.308
so
= arctan(0.308) = 17.1°

From the image you can see that
tan() = 4/13 = 0.308

so
= arctan(0.308) = 17.1°

As the system is at equilibrium, you know that the clockwise torque (caused by the weight of the boom and crate) must be equal in magnitude to the counterclockwise torque (caused by the tension in the cable)
So:
m×g×cos()×L/2 + m×g×cos()×L = T×sin( + )×L
(m/2 + m)×g×cos() = T×sin( + )
T = (m/2 + m)×g×cos() / sin( + )

T = [(179.5 kg)/2 + (89.9 kg)]×(9.81 m/s²)×cos(17.1°) / sin((17.1°) + (17.1°))
T = 2996.81 N

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