Question

You are a summer intern for an architectural firm. An 8.00-m-long uniform steel rod is to be attached to a wall by a frictionless hinge at one end. The rod is to be held at 24.0 ? below the horizontal by a light cable that is attached to the end of the rod opposite the hinge. The cable makes an angle of 35.0 ? with the rod and is attached to the wall at a point above the hinge. The cable will break if its tension exceeds 700 N .

Answer 1

a.) For the sting to break, the tension in it must exceed 700 N

For the given system we balance the net torque acting about the hinge due to the weigh of the rod and the tension in the string.

Now, the net force has to be zero for it to be balanced right before the moment the string snaps.

That is, 700*8* sin35 = m*9.81*4*cos24

Hence, m = 3212.028 / 0.914*9.81*4 = 89.56 kgs

b.) Here, we determine the tension in the string first by balancing the moment about the hinge.

Hence 79.56 * 9.81 * 4 * cos24 = T * 8 * sin35

That is, T = 621 N

Now, we balance the horixontal and vertical forces separately so as to determine the net force by hinge.

That is Fy + Tsin59 = Mg; Hence, Fy = 79.56*9.81 - 621*sin59 = 780.48 - 532.3 = 248.18 N

Again forces along the x axis;

Fx = Tcos59 = 621 * cos 59 = 319.84 N

Therefore the net magnitude of the force being exerted by the rod = sqrt[248.18^2 + 319.84^2] = 404.83 N

c.) For angle above the normal tanθ = Fy/Fx

θ = tanh248.18/319.84 = 0.65 = 37 degrees