A store\'s sign with a mass of 20.0Kg and 3.00 m long, has its center of gravity
ID: 1289864 • Letter: A
Question
A store's sign with a mass of 20.0Kg and 3.00 m long, has its center of gravity at the center of the sign. It is supported by a loose bolt attached to the wall at one end and by a wire at the other end, as shown in the figure. The wire makes an angle of 25 with the horizontal. What is the tension in the wire?
A store's sign with a mass of 20.0Kg and 3.00 m long, has its center of gravity at the center of the sign. It is supported by a loose bolt attached to the wall at one end and by a wire at the other end, as shown in the figure. The wire makes an angle of 25½ with the horizontal. What is the tension in the wire? Please show me how to solve this problem!!!Explanation / Answer
sum torque about the bolt
20*g*1.5=T*sin25*3
solve for T
T=20*g*1.5/(sin25*3)
T=232.1 N
for the bolt, sum forces in the x and y
y
fy=20*g-T*sin25
98.1 N
x
fx=T*cos25
210.4 N
The resultant force is
sqrt(fx^2+fy^2)
232.1 N
(It's equal to the tension in the wire because of the symmetry of the system)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.