# 2.004 MODELING DYNAMICS AND CONTROL II Problem 1

```2.004 MODELING DYNAMICS AND CONTROL II
Spring 2003
Solutions to Problem Set No. 4
Problem 1 This problem can be solved by using the equation to get the velocity at any
point in the 2D rigid body. The equation in this problem is
vA = vA + w &acute; r
Because the ladder slides down, the direction of the angular velocity is counter clockwise
normal to the out of the face. Once we set the magnitude of the angular velocity as w, the
direction of the cross product of w and r can be shown in the Figure 1. However, we
don’t know the magnitude of the cross product yet. Also, we know the direction of
velocity of point A due to the geometric constraint between the point A and the vertical
wall. It’s the vertical direction as shown in the Figure 1.
vB
rw
vA
20&deg;
r
70&deg;
vB
Figure 1
From the triangle includes vA and vB with angle 20 degree as shown in Figure 1,
v A = v B tan(20&deg;) = 2 tan(20&deg;) = 0.73(m / s)
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