Two blocks (one of which is on a ramp) are attached via a string looped over a pulley. The block on the incline has a mass of 5.0 kg; the block hanging downward (suspended by the pulley) has a mass of 7.0 kg. Assume the usual things (non-stretchy massless string, frictionless lightweight pully, air resistance / friction is negligible, etc.) What angle of incline (measured against the horizontal) allows the blocks to have constant velocity?
46
To find the angle of incline that allows the blocks to have a constant velocity, we can start by analyzing the forces acting on the system. The key is to consider the forces on each block separately and then relate them using Newton's second law.
Let's break down the forces on each block:
1. Block on the incline (mass m1 = 5.0 kg):
- Weight (mg1): This force acts vertically downward and is equal to m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- Normal force (N1): This force acts perpendicular to the incline and prevents the block from sinking into the ramp. It is equal in magnitude but opposite in direction to the component of the weight perpendicular to the incline.
- Friction force (f): This force opposes the block's motion and acts parallel to the incline. Since the velocity is constant, the friction force must be equal in magnitude but opposite in direction to the component of the weight parallel to the incline.
2. Hanging block (mass m2 = 7.0 kg):
- Weight (mg2): This force acts vertically downward and is equal to m2 * g.
Now, let's consider the tension in the string (T). Since the string is massless and non-stretchy, the tension is the same for both blocks.
Applying Newton's second law (F = ma) to each block along the ramp, we can write the following equations:
For the block on the incline:
1. In the perpendicular direction: N1 - mg1 * cosθ = 0. (The normal force cancels out the horizontal component of the weight.)
2. In the parallel direction: T - mg1 * sinθ - f = 0. (The tension overcomes the component of the weight along the incline and the friction force.)
For the hanging block:
3. In the vertical direction: T - mg2 = 0.
Since the system is at constant velocity, the acceleration of both blocks is zero. Combining equations 1, 2, and 3, we can solve for the angle of incline (θ) that satisfies these conditions.
First, rearrange equation 2 to solve for the friction force:
f = T - mg1 * sinθ.
Substitute equation 1 into equation 2 to eliminate N1:
T - mg1 * sinθ - (T - mg1 * cosθ) = 0,
T - T + mg1 * sinθ - mg1 * cosθ = 0,
mg1 * sinθ - mg1 * cosθ = 0,
sinθ - cosθ = 0.
Now, divide both sides of the equation by cosθ:
tanθ = 1,
θ = tan^(-1)(1) = 45 degrees.
Therefore, the angle of incline (measured against the horizontal) that allows the blocks to have a constant velocity is 45 degrees.