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On the diagram below, M1g =25.0 and theta = 33.0°. Determine the minimum weight M2 must be in order to accelerate M1 up the plane.
i'd appreciate it if someone could answer this and provide the steps necessary to solve

On the diagram below M1g 250 and theta 330 Determine the minimum weight M2 must be in order to accelerate M1 up the plane id appreciate it if someone could answ class=


Answer :

MathPhys

Answer:

13.6 N

Explanation:

Using the provided free-body diagram, we can apply Newton's second law of motion to balance the forces on each mass. Since no coefficient of friction is given, we will assume the plane is frictionless. If we assume the pulley is also frictionless, then the tension forces are equal, so T₁ = T₂ = T. In the diagram:

  • C is the weight of M₁, C = M₁g. This weight force has components parallel to the plane (A) and perpendicular to the plane (B)
  • D is the normal force, D = N
  • E is the weight of M₂, E = M₂g

Summing the forces on M₂ in the vertical direction:

∑F = ma

T − M₂g = 0

T = M₂g

Summing the forces on M₁ in the parallel direction (use trigonometry to show that A = M₁g sin θ):

T − M₁g sin θ = 0

T = M₁g sin θ

Setting the equations equal:

M₂g = M₁g sin θ

Plug in values:

M₂g = 25.0 N sin 33.0°

M₂g = 13.6 N

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