60.0° 40.0° T2 T3 F. g A bag of cement weighing 325 N hangs in equilibrium from three wires as suggested in the figure. Assuming that the system in equilibrium, what are the tensions T1, T2, and T3 in the wires? 165 N, 253 N, and 325 N. 165 N, 165 N, and 335 N. 162.5 N, 162.5 N, and 325 N. 253 N, 253 N, 506 N. 253 N, 165 N, and 325 N. CEMENT

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60.0°
40.0°
T2
T3
F.
g
A bag of cement weighing 325 N hangs in equilibrium from three wires as suggested in the
figure. Assuming that the system in equilibrium, what are the tensions T1, T2, and T3 in the
wires?
165 N, 253 N, and 325 N.
165 N, 165 N, and 335 N.
162.5 N, 162.5 N, and 325 N.
253 N, 253 N, 506 N.
253 N, 165 N, and 325 N.
CEMENT
Transcribed Image Text:60.0° 40.0° T2 T3 F. g A bag of cement weighing 325 N hangs in equilibrium from three wires as suggested in the figure. Assuming that the system in equilibrium, what are the tensions T1, T2, and T3 in the wires? 165 N, 253 N, and 325 N. 165 N, 165 N, and 335 N. 162.5 N, 162.5 N, and 325 N. 253 N, 253 N, 506 N. 253 N, 165 N, and 325 N. CEMENT
Expert Solution
Step 1

To balance the weight Fg at-rest state,

T3=Fg

Given the value of Fg is,

Fg=325 N

Substituting this value,

T3=325 N

The net force acting in the horizontal direction is,

T2cos40-T1cos60=00.766T2-0.5T1=00.5T1=0.766T2T1=1.532T2

The net force acting in the verticle direction is,

T2sin40+T1sin60-Fg=0

Substituting the known values,

T2sin40+T1sin60-325=00.643T2+0.866T1=3250.643T2+0.8661.532T2=3251.97T2=325T2=165 N

Thus, the value of T1 is,

T1=1.532T2=1.532×165=253 N

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