Question

1. A control arm (see figure below) has mass m and center of gravity / center of mass in the point G. The arm is freely hinged at point O. The points A, B and C are the fixings for a spring and a cord that runs over a station's pulley at point D (neglecting the pulley dimensions). The spring is tensioned when the angle θ = 0 and gives
then a force F = k * s, where s is the extension of the spring. If the arm is in equilibrium, the sum of
every moment of the arm to be zero.
Data: m = 150 kg, k = 2.0 kN · m ^ -1, OG = 1.4m, 2.1m = OB, OC = 4.2m, D = (6.0, 2.0) m, g = 9.8ms ^ -2

Answer 1

(A) Explain why this is appropriate to calculate the moments with respect to the point O.
(B) Identify all the forces that provide a torque (m.a.p. O) and expression both forces and ortsvektorer of forces attack points on the component shape.
Hint: Write all the vectorial quantities that amount multiplied by a unit vector.
(C) Set up and solve the equilibrium equation with respect to the tension force in the rope, T.
Express this as a function of angle T (θ).
(D) See T (θ) graphically.
(E) Suppose that the line maximum withstand a load T = 5.0 kN.
At what angle we reach this value for the clamping force?
(F) What is the maximum value of θ if we do not have to take into account the limitations of the rope?

Answer 2

This is linear algebra course

Answer 3

@ganeshie8 , @IrishBoy123 @Michele_Laino

Answer 4

hi! you there?

Answer 5

nvm

a) why appropriate? well, you can take moments about anywhere you like as long as you do it right. O seems appropriate cos you have a bunch of measurements that are focussed around O. and it's in equilibrium about there [and everywhere else] so you can say that the sum of the moments/torques = 0

have you tried doing the rest yet?

Answer 6

Im trying but I cant figure it out!! Im just trying to set up c = OC* sin (theta) * (1,0) + OC*cos(theta)*(0,1) and so on.. @IrishBoy123

Answer 7

yeah, if you're doing lin alg maybe you should be looking at it as \(\tau = \sum \vec r_i \times \vec F_i\)

so here it's something like, with ccw +ve, \(- \vec{ OG} \times <0, -mg> - \vec{ OB} \times kx ~ \hat{ BA} + \vec{ OC} \times T ~ \hat{ CD} = 0\)

i'd check that but i think that's a pretty mechanical way to go about it

Answer 8

oh, i didn't see the data at the bottom

that means you can write \(\vec {OG} = <1.4 \sin \theta, -1.4 \cos \theta >\)

Answer 9

dunno if that helps ;-|

Answer 10

@IrishBoy123 I have done the tasks now, and I think that e) is 95.8 degrees and f) is 108 degrees