2. In preceding problem, if the mass is 1,356 kg, what is the tension in the sec
ID: 1419284 • Letter: 2
Question
2. In preceding problem, if the mass is 1,356 kg, what is the tension in the second wire to the nearest newton?
3. A 80 kg student sits on a chair which is solely supported by a solid 0.72 meter-long steel rod 0.75 cm in diameter. To the nearest micron (millionth of a meter), what is the change in length of the rod produced by the student's weight? (The Young's modulus for the steel is 20 x 10^10 N/m^2)
5. In the previous problem, if the bar is 1.7 meters long and has a mass is 100 kg, what is the magnitude (to the nearest newton) of the hinge force?
7. In the previous problem, if the force on the left support is 220 newtons and the hanging mass is 25 kg, what is the force on the right support?
8. In Problem 6, if the hanging mass is 23 kg, what is the minimum value of L for the configuration to remain stable to the nearest hundreth of a meter?
9. In Problem 6, what is the maximum value of the hanging mass to the nearest tenth of a kilogram, if L = 1.54 meters and the configuration is to remain stable?
Explanation / Answer
1)
let T1 and T2 are the tensions in the wires 1 and 2.
the points which connectes the three wires is in equilibrium. so, net force acting on the point must be zero.
Apply, Fnetx = 0
T2*cos(60) - T1*cos(30) = 0
T2 = T1*cos(30)/cos(60)
T2 = T1*1732 -------(1)
Apply, Fnety = 0
T1*sin(30) + T2*sin(60) - m*g = 0
T1*0.5 + T1*1.732*sin(60) = m*g
2*T1 = m*g
T1 = m*g/2
= 900*9.8/2
= 4410 N <<<<<<--------------Answer
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.