2) Consider the hanging sign shown here (mass of the sign is 5 kg, length of the horizontal bar = 5 meters, with a tension cable connected at the 4 meter mark – the bar has a mass of 2 kg). a) Calculate the tension in the cable (in N). b) Calculate the horizontal force (and direction) at the left connection point (the bolt/hinge). c) Calculate the vertical force (and direction) at the left connection point (the bolt/hinge). 5.0 m 3.0 m 1.0 m 4.0 m GAS ml = 2 kg a) pivot at left end: 0 = tan = 36.87° T sin 0(4m)= (2)(9.8)(2.5m)+(5)(9.8)(5) Tsin0= 73.5N T=122.5N =T cos 0 = (122.5)cos 36.87 =98N b) Fwall m2 = 5 kg c) F, +T sin 0 = m,g+m,g F = (7)(9.8)– 73.5=-4.9N (this means the vertical force is actually pulling down on the left end!)
2) Consider the hanging sign shown here (mass of the sign is 5 kg, length of the horizontal bar = 5 meters, with a tension cable connected at the 4 meter mark – the bar has a mass of 2 kg). a) Calculate the tension in the cable (in N). b) Calculate the horizontal force (and direction) at the left connection point (the bolt/hinge). c) Calculate the vertical force (and direction) at the left connection point (the bolt/hinge). 5.0 m 3.0 m 1.0 m 4.0 m GAS ml = 2 kg a) pivot at left end: 0 = tan = 36.87° T sin 0(4m)= (2)(9.8)(2.5m)+(5)(9.8)(5) Tsin0= 73.5N T=122.5N =T cos 0 = (122.5)cos 36.87 =98N b) Fwall m2 = 5 kg c) F, +T sin 0 = m,g+m,g F = (7)(9.8)– 73.5=-4.9N (this means the vertical force is actually pulling down on the left end!)
Physics for Scientists and Engineers
10th Edition
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter12: Static Equilibrium And Elasticity
Section: Chapter Questions
Problem 50CP: In the What If? section of Example 12.2, let d represent the distance in meters between the person...
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The answer is already shown, I am just confused about the steps. mostly confused about part a. explain how to use the formula and gets the answers of 73.5 and 122.5. and explain the rest if you can but part is the main focus. thank you.
![2) Consider the hanging sign shown here (mass of the sign is 5
kg, length of the horizontal bar = 5 meters, with a tension
cable connected at the 4 meter mark – the bar has a mass of 2
kg).
a) Calculate the tension in the cable (in N).
b) Calculate the horizontal force (and direction) at the left
connection point (the bolt/hinge).
c) Calculate the vertical force (and direction) at the left
connection point (the bolt/hinge).
5.0 m
3.0 m
4.0m
1.0 m
-36.87
a) pivot at left end :
GAS
0 = tan
= 36.87°
ml = 2 kg
T sin 0(4m)=(2)(9.8)(2.5m)+(5)(9.8)(5) Tsin0 = 73.5N T=122.5N
b) Fa =T cos 0 = (122.5) cos 36.87 = 98N
m2 = 5 kg
wall
c) F, +T sin 0 = m,g+m,g F = (7)(9.8) – 73.5 =-4.9N (this
%3D
means the vertical force is actually pulling down on the left end!)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b50f635-ef3d-4bc5-b774-3501321b612e%2F19bb493d-10c5-46bd-8def-79081625909f%2Fqobrnxa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2) Consider the hanging sign shown here (mass of the sign is 5
kg, length of the horizontal bar = 5 meters, with a tension
cable connected at the 4 meter mark – the bar has a mass of 2
kg).
a) Calculate the tension in the cable (in N).
b) Calculate the horizontal force (and direction) at the left
connection point (the bolt/hinge).
c) Calculate the vertical force (and direction) at the left
connection point (the bolt/hinge).
5.0 m
3.0 m
4.0m
1.0 m
-36.87
a) pivot at left end :
GAS
0 = tan
= 36.87°
ml = 2 kg
T sin 0(4m)=(2)(9.8)(2.5m)+(5)(9.8)(5) Tsin0 = 73.5N T=122.5N
b) Fa =T cos 0 = (122.5) cos 36.87 = 98N
m2 = 5 kg
wall
c) F, +T sin 0 = m,g+m,g F = (7)(9.8) – 73.5 =-4.9N (this
%3D
means the vertical force is actually pulling down on the left end!)
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