Need Holp? 11 5 P076 Jamal and Dayo are lifting a large chest, weighing 325 Ib,

Question

Need Holp? 11 5 P076 Jamal and Dayo are lifting a large chest, weighing 325 Ib, by using the two rope handles attached to either side. As they lift and hold it up so that it is motionless, each handle makes a different angle with respect to the vertical side of the chest (see figure below). If the angle between Jamal's handle and the vertical side is 25.0 and the angle between Dayo's handle and the vertical side of the chest is 30.09, what are the tensions in cach handle? (Due to the nature of this pcoblem, do not use rounded intermediate values in your calculations-including arswers submitted in WebAssign.) 1141732 1483.25 O ponse differs from the correct answer by more than 10%. Double check your calculations. N Need Help? I 10. 15 POB1.M e, accelerating from 94.0 to 132 m/s in 7.60 s. The pilot has hung a 0.180 kg e end of a string from the hinge on the aircraft's canopy. During the dive, the string of the keepsake is seen to remain perpendicular to the top of the canopy as shown in the figure 7:35 PM 2/21/2018 2 ^ Type here to search

Explanation / Answer

Jamal's force is F1; Dayo's force is F2
Assume Jamal is on left side and Dayo is on right side.

Horizontal analysis:

Define Dayo's side (to the right) as the positive direction.

Forces right - forces left = 0
F2 cos(30.0°) - F1 cos (25.0°) = 0

F2=(F1.cos25)/cos30

=1.0465F1.........(1)

Evaluate the trig functions and rearrange the equation to isolate F2.

Vertical analysis:

Forces up - forces down = 0

F1 sin(25.0°) + F2 sin(30.0°) - 325 = 0

sunstitute equation 1 here.

F1 sin(25.0°) + (1.0465F1) sin(30.0°) - 325 = 0

By solving we get F1=FrJ=343.606N

substitute F1 in equation 1 to get F2

and F2=FrD=359.58 N

Now, from your horizontal analysis, you have isolated F2. Substitute the expression for F2 for F2 into your vertical analysis equation. You can now solve for F1 and use that value in your first equation to solve for F2.

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