Dynamics Rigid Bodies Help!!!?

Question

I cannot figure out this problem in my dynamics book: The question states "Points A and B of the 2-m bar slide on the plane surfaces. Point B is moving to the right at 3 m/s. What is the velocity of the midpoint G of the bar?" There is a picture shown of a bar (much like a ladder) that is leaning against a wall. Point A is the top and B is the bottom. Also the bottom of the bar (near point B) makes a 70 degrees angle with the floor. Also a hint is given to use the relative velocity equation: Va= Vb + w x Ra/b. I'm very lost on how to compute the vectors.

Answer 1

I'm assuming that point A of the bar is in contact with a vertical surface (e.g. a wall), and point B of the bar is in contact with a horizontal surface (e.g. a floor). This means that point A of the bar can only move up and down, and point B of the bar can only move left and right.

I'm going to represent the vectors in terms of i, j, and k, which correspond to the x, y, and z components of the vectors, respectively.

First, we have to determine the angular velocity, ω, of the bar. To do so, I'm going to make use of the following equation to take advantage of the fact that the velocity of point A can only be in the y direction, (i.e. (Va) i = 0):

Va = (Va) j

= Vb + ω x Ra/b

= 3i + {ωk x [2cos(70°) i + 2sin(70°) j] }

Recall:

i x j = k, j x k = i, k x i = j, k x j = -i, etc.

=> (Va) j

= 3i + [2ωcos(70°) j - 2ωsin(70°) i]

= [3 - 2ωsin(70°)] i + [2ωcos(70°)] j

Equate the left and right hand sides of the equation to obtain:

[2ωcos(70°)] j = (Va) j, and

3 - 2ωsin(70°) = 0

=> ω = 3 / [2sin(70°)] = 1.596 rad/s

So the angular velocity of the rod, ω = 1.596 rad/s.

We can now solve for the velocity of the midpoint G, Vg, of the bar, where,

Vg = (Vg) i + (Vg) j

= Vb + ω x Rg/b

= 3i + {ωk x [1cos(70°) i + 1sin(70°) j] }

= 3i + [ωcos(70°) j - ωsin(70°) i]

= [3 - ωsin(70°)] i + [ωcos(70°)] j

Equate the left and right hand sides of the equation to obtain:

(Vg) i = [3 - ωsin(70°)] i

(Vg) j = [ωcos(70°)] j

We know that the angular velocity of the rod, ω = 1.596 rad/s

So just plug in ω into the above equations to obtain:

(Vg) i = [1.50 m/s] i

(Vg) j = [0.546 m/s] j

Therefore, the velocity of point G,

Vg = (Vg) i + (Vg) j = [1.50 i + 0.546 j ] m/s