A block with mass m1 on a plane inclined at angle ƒÑ to the horizontal is connected to a second hanging block with mass m2 by a cord passing over a small, frictionless pulley, as shown. The coefficient of static friction is Us ƒÝ
s and the coefficient of kinetic friction is UkƒÝ
k. Find the mass m2 for which block m1, moves up the plane at constant speed once it is set in motion.
Nice one UCT student ^^
To find the mass m2 for which block m1 moves up the plane at a constant speed, we need to consider the forces acting on both blocks.
Let's analyze the forces acting on block m1:
1. The weight of m1 acts vertically downwards and can be calculated as W1 = m1 * g, where g is the acceleration due to gravity.
2. The normal force acting perpendicular to the incline opposes the weight and can be calculated as N1 = m1 * g * cos(Ą), where cos(Ą) is the cosine of the angle of inclination.
3. The force of static friction opposes the motion of block m1, and its magnitude is given by fs = Us * N1, where Us is the coefficient of static friction.
The forces acting on block m2 are as follows:
1. The weight of m2 acts vertically downwards and can be calculated as W2 = m2 * g.
2. The tension in the cord, T, pulls block m2 upwards.
Since both blocks are connected by a cord passing over a small, frictionless pulley, the tension in the cord is the same for both blocks.
Now, the key to finding the mass m2 for which block m1 moves up the plane at a constant speed is to understand that the force of static friction acts as the limiting factor to the motion of block m1. Once it starts moving, we assume the force of static friction changes to the force of kinetic friction, which is given by fk = Uk * N1, where Uk is the coefficient of kinetic friction.
To find the mass m2, we need to consider the balance of forces:
On block m1:
- The force component pulling m1 up the incline is T * sin(Ą).
- The force component opposing the motion down the incline is -T * cos(Ą) + fs.
Since m1 is moving up the plane at a constant speed, the net force on m1 in the horizontal direction is zero. Therefore, we have:
T * sin(Ą) = T * cos(Ą) - fs
Substituting fs = Us * N1 and N1 = m1 * g * cos(Ą), we get:
T * sin(Ą) = T * cos(Ą) - Us * m1 * g * cos(Ą)
We can simplify this equation to:
T = Us * m1 * g * cos(Ą) / (sin(Ą) + cos(Ą))
Now, let's consider the forces acting on block m2:
- The weight of m2 is pulling it downwards with a force of m2 * g.
- The tension in the cord, T, is pulling m2 upwards.
Since m2 is assumed to be moving upwards at a constant speed, the net force on m2 in the vertical direction is zero. Therefore, we have:
T = m2 * g
Combining this equation with the equation we derived for T, we can solve for m2:
m2 * g = Us * m1 * g * cos(Ą) / (sin(Ą) + cos(Ą))
Simplifying this equation, we get:
m2 = (Us * m1 * cos(Ą)) / (sin(Ą) + cos(Ą))
Therefore, the mass m2 for which block m1 moves up the plane at a constant speed is given by:
m2 = (Us * m1 * cos(Ą)) / (sin(Ą) + cos(Ą))