Question

(14 points) The figure below shows the bird's eye view of three forces acting on a long, thin rod with a length of 2.40 m that can pivot about a point one-fourth of its length from one end of the rod. The moment of inertia of the rod about its pivot point is 37.8 kg.m. Forces 1 and 2 each have a magnitude of 120 N. Force 1 acts perpendicularly to the rod at the end of the rod nearest to the pivot and force 2 acts midway between that end of the rod and the pivot, making an angle of 0 = 60.0° with the rod. Force 3 has a magnitude of 200 N and also acts perpendicularly to the rod. (Icm = ML for a uniform long, thin rod) F1 F2 (a) What is the mass of the beam? (b) What is the distance d that force 3 acts from the pivot if the angular acceleration of the rod is 4.00 rad/sand points into the page?

Answer 1

(a) Given the length of the rod is L = 2.40m and the moment of inertia of the rod about the pivot point is I = 37.8kgm2.

The centre of mass of the rod is at L/2 diatance from the end of the rod. The moment of inertia about the centre of mass of the rod is given as,

12

The pivot point is L/4 from the one end of the rod. So the distance of the pivot point from the centre of mass is L/2 - L/4 = L/4. The moment of inertia about the pivot point is given by,

$I=I_{cm}+M(\frac{L}{4})^{2}=\frac{ML^{2}}{12}+\frac{ML^{2}}{16}=\frac{7ML^{2}}{48}$

Given,

$\frac{7ML^{2}}{48}=37..8kgm^{2}$

$M=\frac{37.8\times 48}{7L^{2}}=\frac{37.8\times 48}{7\times (2.40)^{2}}=45kg$

So the mass of the rod (beam) is 45kg.

(b) We know that the torque acting on the system is given by

$\tau=F\times R = I\alpha$

where F is the force acting on the rod, R is the perpendicular distance between the line of action of the force and the pivot point, I is the moment of inertia of the rod about the pivot point and $\alpha$ is the angular acceleration of the rod.

Given $\alpha$ = 4 rad/s2 and points into the page. So the rod is rotating in clockwise direction. Therefore clockwise torque is greater than anticlockwise torque. Therefore

$\tau_{clockwise}-\tau_{anti-clockwise}=I\alpha$

F3 is at a distance d from the pivot point, F1 is at a distance L/4 from the pivot point. F2 can be resolved into sin and cos components. Only sin component can rotate the rod and it is at a distance L/4 from the pivot point. So,

$F_{3}\times d -(F_{1}\times \frac{L}{4}+F_{2}sin\theta \times \frac{L}{4})=I\alpha$

Given F3 = 200N, F1 = F2 = 120N and $\theta$ = 60$\degree$. Therefore,

$200d-120\times\frac{2.40}{4}-120sin(60\degree)\times \frac{2.40}{4}=37.8\times 4$

$200d-72-62.35=151.20$

$200d=285.55$

$d=1.43m$

So the distance d from F3 to pivot is 1.43m.