Question

a 50.0 kg 6.00 m awning is held up by a wire in its middle. A 15.0 kg sign hangs at the end. Calculate the tension in the massless wire and the magnitude of the net force from the wall acting on the awning. The angle between the awning and the wire in 40 degrees

Answer 1

Given that

mass m=15 kg

mass M=50 kg

angle=40 degree

length L=6 m

now we find the tension in the massless wire

Sum of Torque=0=(mg)(L)+(Mg)(L/2)-(T)(L/2)Sin40)

L's cancel

(mg)+(Mg/2)=Tsin40/2

(15*9.8)+(50*9.8/2)=T*0.643

147+245=0.643*T

tension T=609.6 N

now we find the net force act on the wall

the net force F=T*sin(theta)=609.6*sin40=391.8 N