Question

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1) draw the forces acting on each mass *pulleys are "ideal". (use appropriate coordinate systems)

2) sum the forces in x,y and find acceleration

Free Body Diagram. Three masses mi, m2, and m3 are attached by a light rope that does not stretch. The rope goes over two massless and frictionless pulleys. Draw the complete free body diagram on each of the three masses. The coefficient of kinetic friction between mi and the incline is plk. All other sur faces are frictionless. Apply Newton's Second Law and find the acceleration of the masses. When c. and ß become 90 degrees, what does the acceleration reduce to?

Answer 1

Date: Fage. Solution Free body diagrams of each block - a Nz 0 % mg o3 misid migoosa megsing Ne = normal force on mi due to surface mig- mass weight of mi fr= friction force on mi due to surface T = Tension force due to pulley between my and mig= weight of m2 N₂ = normal force on M₂ To = Tension force due to and mz due to surface pulley betwan ma mig = no o Weight of me Dormal force on my due to surface Nows equations of motion for each block -

amigcosa migsina et fr= Mi Ni. 1) Ex = Ti - miosina - MkNe = mia { Fy = Ni - miglout o. .. Ni= m. geosa Pulting in (i), we get To - migsina - Memigcosa = mia For mai- (ii) m2 — (IT) Eb= To - Ti = mga Efy = N₂-mag=0.

3 Efx = magsin ß - T2 = mza - Ciu) Efy = Nz - magrosd = 0 Now, adding equations (i), (N) and (iv) we get mugsinß - migsina - Mamigcosa = miat mama

Date: Page . a = masinß - misina - Mimicosa mit 2+ M3 al - our equation for Now, a = ß= 90° acceleration reduces so to a= masin go - mising o- Mimicos go mitm+m3 - ma-mi-o singo = 1 and cos go= 0] mitmet my mitm₂ + m3