The Figure shows two blocks, block 1 and block 2, held in equilibrium by a massless, unstretchable string. The mass of block 1 is 14.2 kg, the mass of block 2 is 6.2 kg, and the angle theta is 26o
The system begins to slip if any additional mass is added to block 2. What is the coefficient of static friction between block 1 and the plane on which it rests?
To solve this problem, we need to consider the forces acting on block 1. In the given system, the tension in the string pulls block 1 while the weight of block 1 pulls it downwards. Additionally, there is a force of static friction between block 1 and the plane that prevents it from slipping.
Let's break down the forces acting on block 1:
1. Tension force (T): The tension in the string acts horizontally, pulling block 1 towards the right.
2. Weight force (W1): The weight of block 1 acts vertically downward. The magnitude of the weight force is given by W1 = m1 * g, where m1 is the mass of block 1 and g is the acceleration due to gravity (approximately 9.8 m/s^2).
3. Normal force (N1): The normal force acts perpendicular to the plane and counteracts the weight force. It can be calculated as N1 = m1 * g * cos(theta), where theta is the angle between the plane and the horizontal.
4. Friction force (f1): The static friction force acts parallel to the plane and counteracts the tension force. It prevents block 1 from slipping. The magnitude of the static friction force can be calculated as f1 = u * N1, where u is the coefficient of static friction.
In equilibrium, the sum of the horizontal forces is zero (T - f1 = 0) and the sum of the vertical forces is zero (N1 - W1 = 0).
Now let's solve for the coefficient of static friction:
1. From the horizontal force equation (T - f1 = 0), we can substitute T = m2 * g * sin(theta), where m2 is the mass of block 2. We use sin(theta) because the tension in the string is divided between the horizontal and vertical components.
2. Substitute N1 = m1 * g * cos(theta) and W1 = m1 * g into the vertical force equation (N1 - W1 = 0) to get N1 = W1 = m1 * g.
3. Substitute the values we have: T = 6.2 kg * 9.8 m/s^2 * sin(26°), N1 = 14.2 kg * 9.8 m/s^2, and f1 = u * N1.
4. Rearrange the horizontal force equation to solve for the coefficient of static friction: u = T / N1.
5. Calculate the value for the coefficient of static friction by substituting the values we have: u = (6.2 kg * 9.8 m/s^2 * sin(26°)) / (14.2 kg * 9.8 m/s^2).
6. Simplify the expression: u = (6.2 * sin(26°)) / 14.2.
7. Calculate the numerical value: u ≈ 0.170.
Therefore, the coefficient of static friction between block 1 and the plane is approximately 0.170.