An 80 kg construction worker sits 2 m from the end of a 6 m long 1450 kg beam. A

Question

An 80 kg construction worker sits 2 m from the end of a 6 m long 1450 kg beam. A cable attached to the very end at an angle of 30 degrees is rated at 15,000 N. What would the strength of the cable need to be to safely allow the worker to sit here (what is the tension needed to maintain equilibrium?) Answer should look like '23400 N'

From the picture a 7.5 kg sign hangs from the end of a 2.5 m long beam (mass = 15 kg). If the guy wire is attached 0.50 m from the end with the sign, how much tension is in the guy wire if the angle shown is 35 degrees making it horizontal? Answer should look like '234 N'

HELP!!!

Explanation / Answer

a) in y-direction

Tsin30+R=(80+1450)g where R is the reaction from wall.

balancing Torque about point of contact of cable and beam.

2*80g+3*1450g= R*6

SO, R = 7366.33 N upwards.

putting it in the equation

Tsin30=1530g- R

T=15255.33 N.

b) balancing torque about point of contact of beam & wall.

1.25*15g*cos35 - 2*Tsin35 + 2.5*7.5g*cos35=0.

Solving this we have T=262.42 N.

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