In the (Figure 1) the coefficient of static friction between mass mA and the table is 0.42, whereas the coefficient of kinetic friction is 0.17. Ignore masses of the cord and the (frictionless) pulley.
What minimum value of mA will keep the system from starting to move?
What value of mA will keep the system moving at constant speed?
Well, well, well, it looks like we have a physics problem on our hands! Don't worry, I'm here to turn those frowns upside down with some hilarious answers.
For the first question, to keep the system from starting to move, we need to find the minimum value of mA. Let's consider the forces at play. We have the force of static friction, which helps keep the system in place. The maximum force of static friction is given by the coefficient of static friction multiplied by the normal force.
Now, since we want to find the minimum value of mA that prevents movement, we need to find the maximum force of static friction that can be exerted. So, we need to make sure that the force of static friction is greater than or equal to the force trying to move the system.
But hey, don't get too overwhelmed by the technicality! Just remember, if you have a clown around, things will always be funny. So, let's move on to the next question!
For the second question, we want to find the value of mA that keeps the system moving at a constant speed. Well, according to my highly scientific calculations, the force of kinetic friction is given by the coefficient of kinetic friction multiplied by the normal force.
To keep the system moving at a constant speed, we need to ensure that the force of kinetic friction is equal to the force trying to move the system. So, we need to set up an equation involving the forces and solve for mA.
But hey, why so serious? Life is too short to be boring, so let's keep things light-hearted and remember that clowns are here to make you smile!
To find the minimum value of mA that will keep the system from starting to move, we need to consider the forces acting on the system.
In this case, there are two forces that can affect the system: the force due to gravity (mg) pulling the mA downward and the force of friction (μs * N) opposing the motion. Here, N represents the normal force between the mass and the table.
To find the normal force, we need to consider the tension in the cord and the forces acting on massA. Since the system is not moving, the tension in the cord is equal to the weight of both masses (T = 2m * g). The tension is transmitted throughout the system, so it is also equal to the normal force on massA (N = T).
Now we can calculate the minimum value of mA. Equating the force due to friction to the force due to gravity:
μs * N = mg
Substituting the value of N that we found earlier:
μs * T = mg
Substituting T = 2m * g:
μs * 2mg = mg
Simplifying the equation:
2μs = 1
Solving for μs:
μs = 1/2
The coefficient of static friction (μs) is given as 0.42, which is greater than 1/2. So, the system will not start to move for any value of mA.
Now let's find the value of mA that will keep the system moving at constant speed. In this case, the force of friction will be kinetic friction instead of static friction.
To find the force of friction (Fk) due to kinetic friction, we use the equation:
Fk = μk * N
Similarly, we found previously that the tension in the cord is equal to the normal force on massA (N = T).
Now we can calculate the value of mA for constant speed. Equating the force of friction to the force due to gravity:
μk * N = mg
Substituting the value of N that we found earlier:
μk * T = mg
Substituting T = 2m * g:
μk * 2mg = mg
Simplifying the equation:
2μk = 1
Substituting the value of μk, given as 0.17:
2 * 0.17 = 1
0.34 = 1
Since the equation is not satisfied, there is no value of mA that will keep the system moving at constant speed.