the mass M = 1 kg rests on a horizontal plane. Calculate the maximum value of the angle θ0 between the mass m and the vertical axis, that does not produce the displacement of mass M.

Question

the mass M = 1 kg rests on a horizontal plane. The coefficient of static friction relative to the contact between the body and the plane is

μs = 0.5. An inextensible wire of negligible mass, connects the mass M to the mass

m = 0.3 kg, suspended in vacuum. The pulley C rotates without friction and has negligible mass. Calculate the maximum value of the angle θ0 between the mass m and the vertical axis, that does not produce the displacement of mass M.

Hint: Notice that the tension in the wire cannot be neglected.

Answer 1

Given:

1. Mass M=1 kgM = 1 \, \text{kg}M=1kg (on the horizontal surface)
2. Mass m=0.3 kgm = 0.3 \, \text{kg}m=0.3kg (hanging at an angle θ₀ to the vertical)
3. Coefficient of static friction μs=0.5\mu_s = 0.5μs​=0.5
4. Pulley is frictionless and massless
5. Wire is inextensible and massless

We assume gravity g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2

Step 1: Forces Acting on Each Mass

On mass M (on table):

1. Weight: WM=Mg=9.8 NW_M = M g = 9.8 \, \text{N}WM​=Mg=9.8N
2. Normal force N=WMN = W_MN=WM​ (since there's no vertical force from the wire)
3. Friction force (max static): fmax=μsN=0.5⋅9.8=4.9 Nf_{\text{max}} = \mu_s N = 0.5 \cdot 9.8 = 4.9 \, \text{N}fmax​=μs​N=0.5⋅9.8=4.9N
4. Horizontal tension from the wire: Tcos⁡θ0T \cos \theta_0Tcosθ0​

On mass m (hanging):

1. Weight: Wm=mg=0.3⋅9.8=2.94 NW_m = m g = 0.3 \cdot 9.8 = 2.94 \, \text{N}Wm​=mg=0.3⋅9.8=2.94N
2. Tension in wire TTT, angled from the vertical by θ0\theta_0θ0​
3. At equilibrium: Vertical component of tension balances weight:
4. Tsin⁡θ0=Wm=2.94⇒T=2.94sin⁡θ0T \sin \theta_0 = W_m = 2.94 \Rightarrow T = \frac{2.94}{\sin \theta_0}Tsinθ0​=Wm​=2.94⇒T=sinθ0​2.94​

Back to mass M:

For mass M to not move, the horizontal force (from the wire tension) must be less than or equal to the max static friction:

Tcos⁡θ0≤fmax=4.9T \cos \theta_0 \leq f_{\text{max}} = 4.9Tcosθ0​≤fmax​=4.9Substitute T=2.94sin⁡θ0T = \frac{2.94}{\sin \theta_0}T=sinθ0​2.94​:

Final Calculation:

θ0≥cot⁡−1(1.6667)≈tan⁡−1(0.6)≈30.96∘\theta_0 \geq \cot^{-1}(1.6667) \approx \tan^{-1}(0.6) \approx 30.96^\circθ0​≥cot−1(1.6667)≈tan−1(0.6)≈30.96∘ ✅ Answer:

The maximum angle θ₀ that does not cause M to move is:

30.96∘\boxed{30.96^\circ}30.96∘

​Any angle greater than or equal to this keeps M at rest.