My Notes 0/5 points | Previous Answers A rigid, vertical rod with a mass of 2.1

Question

My Notes 0/5 points | Previous Answers A rigid, vertical rod with a mass of 2.1 kg simply rests on the floor and is held in place by static friction. The coefficient of static friction between the rod and the floor is 1/8. The rod also has a wire connected between its top end and the floor, as shown in Figure 11-31. A horizontal force F is applied at the midpoint of the rod 45° Figure 11-31 Find the greatest force F that can be applied at the midpoint of the rod without causing it to slip. 2.5725

Explanation / Answer

As the rod is in equilibrium net force and net torque acting on it must be zero.

let N is the nromal force exerted by the floor on the rod.

Let T is the tension in the rod.

Apply, net torque about bottom of the rod = 0

T*L*sin(45) - (L/2)*F*sin(90) = 0

T*sin(45) = F/2

T = F/(2*sin(45))

T = 0.707*F

now apply, Fnety = 0

N - T*sin(45) - m*g = 0

N = T*sin(45) + m*g

= F/2 + 2.1*9.8

= F/2 + 20.58

now apply, Fnetx = 0

F - f_staticfriction - T*sin(45) = 0

F - N*mue_s - T*sin(45) = 0

F - (F/2 + 20.58)*(1/8) - F/2 = 0

==> F = 5.88 N <<<<<<<<------------------------Answer

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