- #1
lames said:Homework Statement
block1 2kg
block2 2.5kg
coefficient of static friction =.5
coefficient of kinetic friction= .3
A)how can you find the acceleration and tension?
which coefficient do you use?
danielu13 said:For the Tension, you calculate the amount of tension in each section of string and add them together. The tension in T1 is calculated by the force of friction with is
[itex]F_{friction} = F_N\mu_k[/itex]
For this, you use the coefficient of kinetic friction, as I am assuming that the force is being calculated when the blocks are already moving.
So,
[itex]T = T_1 + T_2[/itex]
[itex]T_1 = F_N\mu_k[/itex]
[itex]T_1 = 5.886 N[/itex]
[itex]T_2 = mg[/itex]
[itex]T_2 = 24.525 N[/itex]
[itex]T = 30.411 N[/itex]
For the acceleration you use Newton's Second:
[itex]\Sigma\vec{F} = m\vec{a}[/itex]
The sum of the forces in the system that we are looking at is contained within the tension of the wires, so
[itex]30.411 N = (4.5 kg)a[/itex]
[itex]a = 6.758 \frac{m}{s^2}[/itex]
I disagree entirely with danielu3's analysis and answers.lames said:are you sure those are the right answers? did you look at the picture?
You can't add tensions like that. Consider two equal masses, m, hung over a light pulley by a string. The tension in the string is mg, not 2mg. Now make the masses unequal. What is the tension now?danielu13 said:Yes, the tension in the string is constant. However, there are two components to the tension in the string; one is caused by the force of gravity acting on mass 2 and the other is the force of friction acting on mass 1.
Don't just guess, try to reason it out. Think about the acceleration.danielu13 said:If there were two unequal masses on a pulley would the tension be
[itex]T = g\frac{m_1+m_2}{2}[/itex]
Let the acceleration be a (down for the heavier, up for the lighter) and the tension be T.danielu13 said:Well, based on what you said, I was thinking that the tension is based on the even distribution of the masses, which is why I'm thinking the average of the masses. But I have some feeling that it should be g(m2-m1) since the pulley makes the gravitational forces counteract each other. Am I at least thinking in the right direction right now?
The purpose of solving inclined pulley table problems is to determine the acceleration and tension of an object on an inclined plane connected to a pulley system. This is often used in physics and engineering to understand the motion and forces acting on objects in real-world scenarios.
The acceleration of an object on an inclined pulley table can be calculated using the formula a = (m1 - m2 * sinθ)/(m1 + m2), where m1 is the mass of the object on the inclined plane, m2 is the mass of the object hanging from the pulley, and θ is the angle of the incline.
The tension in the rope of an inclined pulley table can be calculated using the formula T = m2 * g * sinθ, where m2 is the mass of the object hanging from the pulley, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline. This formula assumes a massless, frictionless pulley.
Friction on an inclined pulley table can affect the acceleration and tension by adding an additional force that must be overcome. The coefficient of friction can be included in the equations for acceleration and tension, depending on the specific scenario.
Inclined pulley tables have various real-world applications, such as elevators, cranes, and conveyor belts. They are also used in physics experiments to study motion and forces, and in engineering to design and analyze structures and machines.