In this problem they show a diagram of a pulley system with an inclined plane (25 degree incline). There is a box called m1=47kg resting on the incline plane and m1=35kg hanging on the vertical line. The question reads: The coefficient of friction between m1 and the surfvace of the inclined plane are Ustatic=0.42 and Ukinetic=0.19.
a)if the masses are held in place and then released , will they start to move? ------> I know the answer is NO.
b)how much mass would you have to add to m2 to cause the masses to begin to move? ----> The answer should be 2.8kg but I don't know how the book got this answer.
c)If you said no to a) and added the mass that you calculated in b), what would be the acceleration of the masses? ---> answer given in the back of textbook is 5.7 m/s^2
ANY help with this problem would be greatly appreciated!! THANK YOU SO MUCH in advance :)
the acceleration of the mass will be 7.3m/s2
To solve this problem, let's break it down step by step:
a) To determine if the masses will start to move, we need to compare the force of gravity pulling them down the incline to the force of friction resisting their motion. The force of gravity acting on m1 is given by:
Fg1 = m1 * g
Where m1 is the mass of m1 (47kg) and g is the acceleration due to gravity (9.8 m/s^2). The component of the force of gravity acting down the incline is given by:
Fg1_parallel = Fg1 * sin(25°)
The force of friction opposing the motion of m1 is given by:
Ff1 = μ * N1
Where μ is the coefficient of friction (Ustatic = 0.42), and N1 is the normal force acting on m1. The normal force is equal to the perpendicular component of the force of gravity, which is given by:
N1 = Fg1 * cos(25°)
Now, compare the frictional force to the component of the force of gravity acting down the incline:
Ff1 ≤ Fg1_parallel
Plug in the values:
μ * N1 ≤ Fg1_parallel
(0.42) * (m1 * g * cos(25°)) ≤ (m1 * g * sin(25°))
Simplify and solve for m1:
0.42 * (47kg * 9.8 m/s^2 * cos(25°)) ≤ 47kg * 9.8 m/s^2 * sin(25°)
m1 cancels out, and we are left with:
0.42 * (9.8 m/s^2 * cos(25°)) ≤ 9.8 m/s^2 * sin(25°)
Calculating this inequality, we find that it is false. Therefore, the masses will not start to move.
b) To determine how much mass we need to add to m2 to cause the masses to begin to move, we need to calculate the maximum force of static friction that can be exerted between m1 and the inclined plane. The maximum force of static friction is given by:
Ff1_static_max = μ_static * N1
Where μ_static is the coefficient of static friction (Ustatic = 0.42), and N1 is the normal force acting on m1 (which we have already calculated).
The force required to overcome the static friction is equal to the force of gravity acting down the incline:
Fg1_parallel = Ff1_static_max
(m1 * g * sin(25°)) = μ_static * (m1 * g * cos(25°))
Simplify and solve for m1:
(m1 * g * sin(25°)) / (m1 * g * cos(25°)) = μ_static
sin(25°) / cos(25°) = μ_static
μ_static = tan(25°)
Plug in the value for μ_static and solve for m1:
0.42 = tan(25°)
Now, the force required to move the masses is equal to the combined force of gravity acting down the incline and the additional mass (m2) hanging vertically:
(m1 * g * sin(25°)) + (m2 * g) = μ_kinetic * (m1 * g * cos(25°))
Simplify and solve for m2:
(m2 * g) = μ_kinetic * (m1 * g * cos(25°)) - (m1 * g * sin(25°))
Plug in the values for μ_kinetic, m1, and solve for m2:
m2 = (μ_kinetic * (m1 * g * cos(25°)) - (m1 * g * sin(25°))) / g
m2 = (0.19 * (47kg * 9.8 m/s^2 * cos(25°)) - (47kg * 9.8 m/s^2 * sin(25°))) / 9.8 m/s^2
Calculating this equation, we find that m2 ≈ 2.8kg.
c) To calculate the acceleration of the masses, once the additional mass (m2) has been added, we can use Newton's second law:
ΣF = m_total * a
The net force acting on the system is equal to the force of gravity acting down the incline minus the frictional force:
ΣF = (m2 * g) - (m1 * g * sin(25°))
The total mass is the sum of m1 and m2:
m_total = m1 + m2
Plug in the values for m2 and solve for m_total:
m_total = m1 + 2.8kg
Finally, use Newton's second law to calculate the acceleration (a):
(m2 * g) - (m1 * g * sin(25°)) = (m1 + 2.8kg) * a
Plug in the values for m2, m1, and solve for a:
(2.8kg * 9.8 m/s^2) - (47kg * 9.8 m/s^2 * sin(25°)) = (47kg + 2.8kg) * a
Calculating this equation, we find that a ≈ 5.7 m/s^2.
I hope this helps! Let me know if you have any further questions.