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Linear Impulse/Momentum for Rigid Bodies in a Plane (example continued)

Step 2: Calculate the linear impulse acting on the body.

t = 2 t = 2

∫ F dt = ∫ [ 4t cos θ) i + ( ˗ 4t sin θ + N ˗ mg ) j ] dt

t = 0 t = 0

t = 2 t = 2

∫ F dt = ∫ [ 4t cos θ) i dt = 8 cos θ i lb sec = 4 √3 i lb sec

t = 0 t = 0

since the body is in equilibrium in the y-direction: ˗ 8 sin θ + N ˗ mg = 0

Step 3: Calculate the change in linear momentum of the body.

( m v_Cx2 ˗ m_vCx1) i + ( m v_Cy2 ˗ mv_Cy1) j = ( m v_Cx2 ) i = 2 v_Cx2 i

Note: There is no change in the y˗component of linear momentum since the body is

in equilibrium in the y-direction and mv_Cy1 = 0.

Finally set the linear impulse equal to the change in linear momentum.

4 √3 i = 2 v_Cx2 i

So v₂ = 2 √ 3 i ft/sec (result)