Two identical masses are connected by a (massless, stretchless) string across a (massless, frictionless) pulley. The mass on the right is held so that the string connected to it is horizontal. Everything is initially at rest; the tension is the string is zero. The mass on the right is then released and begins to swing downward. At what angle θwill the mass on the inclined plane begin to slide if the coefficient of static friction between the mass and the plane is 0.50? The indicated angle of the incline is 25.0 O.
mass on the right is horizontal, mass on left is initially horizontal. Where is the incline?
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To determine the angle θ at which the mass on the inclined plane will begin to slide, we can analyze the forces acting on the mass.
1. Gravitational force: The weight of the mass on the inclined plane can be resolved into two components: one parallel to the incline and one perpendicular to the incline. The component parallel to the incline is given by:
F_parallel = m * g * sin(θ)
where m is the mass and g is the acceleration due to gravity.
2. Normal force: The normal force is the force exerted by the inclined plane on the mass and is equal to the perpendicular component of the weight:
F_normal = m * g * cos(θ)
3. Frictional force: The maximum static frictional force is given by:
F_friction = μ_s * F_normal
where μ_s is the coefficient of static friction.
For the mass to begin sliding, the force of static friction must equal the gravitational force component parallel to the incline:
F_friction = F_parallel
Substituting the expressions for F_friction and F_parallel:
μ_s * F_normal = m * g * sin(θ)
Since we are given that the coefficient of static friction is 0.50, we can substitute this value into the equation:
0.50 * (m * g * cos(θ)) = m * g * sin(θ)
Next, we can cancel out the mass and acceleration due to gravity:
0.50 * cos(θ) = sin(θ)
Now, we can solve for θ:
tan(θ) = 0.50 / 1
θ = atan(0.50)
θ ≈ 26.57 degrees
Therefore, the mass on the inclined plane will begin to slide at an angle of approximately 26.57 degrees.
To find the angle at which the mass on the inclined plane will begin to slide, we need to consider the forces acting on it.
First, let's draw a diagram to visualize the setup.
θ
─(m1)───────────┐│
││
││ ┌───────┐
││ │ m2 │
││ └───────┘
││ /
││ /
││ /
││ /
────┘│──/
│ /
(g) │/ ← Direction of gravity
In this diagram, m1 represents the mass on the inclined plane, m2 represents the mass hanging from the string, and θ is the angle of incline.
Now, let's analyze the forces acting on m1. There are two forces to consider:
1. The component of gravity parallel to the inclined plane: mg*sin(θ)
2. The force of static friction between m1 and the inclined plane: fs (unknown)
Since the mass on the right is released, it starts moving downward and the tension in the string increases. As the tension in the string increases, the mass on the inclined plane experiences an additional force in the upward direction. We can call this force as T'.
Let's consider the forces acting in the vertical direction. These forces should balance each other because the mass on the inclined plane is not accelerating in the vertical direction.
∑Fy = 0
mg*cos(θ) - T' = 0
We can express T' in terms of the tension in the string (T) using the relationship:
T' = T*sin(θ)
Combining the two equations:
mg*cos(θ) - T*sin(θ) = 0
Next, let's consider the forces acting in the horizontal direction. Since the mass on the inclined plane is not accelerating in the horizontal direction:
∑Fx = 0
fs - T*cos(θ) = 0
Now, we have two equations with two unknowns (fs and θ). We can solve these equations simultaneously to find the angle θ at which the mass on the inclined plane begins to slide.
1. Start by substituting T*cos(θ) in the second equation with T' from the first equation:
fs - T*cos(θ) = 0
fs = T*sin(θ)
2. Substitute T'*sin(θ) for fs in the second equation:
T*sin(θ) = T*cos(θ)
tan(θ) = 1
θ = arctan(1)
Now, evaluating the arctan(1) using a calculator, we find that θ = 45 degrees.
However, note that the given indicated angle of the incline is 25 degrees. Since θ > indicated angle, the mass on the inclined plane will slide before reaching 25 degrees.