Kinetics - Body Translating in a Plane

Example: (continued) d²x/dt² = 2 d²y/dt² or a_A = 2a_B

Now apply Euler’s 1^st Law: ΣF = ma for Block A → +

T – μW_A = (W_A/g) d²x/dt² = 2(W_A/g) d²y/dt² (1)

ΣF = ma for Block B ↓ +

W_B – 2T = (W_B/g) d²y/dt² (2)

Now W_B = 4 W_A So 4W_A – 2T = (4W_A/g) d²y/dt² (3)

Combine (1) and (3) to eliminate T and solve for d²y/dt² . The result is:

d²y/dt² = [ (2 – μ)/4 ] g

d²x/dt² = 2[ (2 – μ)/4 ] g

Integration gives:

dy/dt = [ (2 – μ)/4 ] g t + C₁

dx/dt = 2[ (2 – μ)/4 ] g t + C₂

Integration gives the velocities for each block. Since both blocks start from rest the

constants of integration will both be zero. So for μ = ¼ and t = 2 seconds

V_A = (7/2)g → ft/sec (result)

and V_B = (7/4)g ↓ ft/sec (result)

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