Motion with Incline: Masses, Acceleration, Tension & Friction | Incline Homework

[jwxie]

Homework Statement

Masses m1 = 4.0kg and m2 = 9.0kg are connected by a light string that passes over a frictionless pulley. As shown in Figure, m1 is held at rest on the floor and m2 rests on a fixed incline of 40 degree. The masses are released from rest, and m2 slides 1.00m down the incline in 4.00s.

Determine a) the acceleration of each mass b) the tension in the string and c) the coefficient of kinetic friction between m2 and the incline.

Homework Equations

Fnet = ma
distance formula
d = 1/2at^2 when vi = 0

The Attempt at a Solution

Well, I know the acceleration for m2 is 0.125m/s^2 using the distance formula when vi=0.
but then i got stuck at the rest, i tried to find acceleration of m1 but didn't work out at all.

well if i tried
m2gcos - m1g = m2a1+m1a2
(since incline plane i can find fg by m2gCos)
but the problem is, i totally ignore the friction Ff i think...
this is why i come up with answer like 6.8m/s^2 for mass1.

the rest will be wrong if i can't set up a system of equation for #1
i tried, like

m2gCos + Ft = m2a
m1g + (-Ft) = m1a

Ft means force of tension

 

[hage567]

You need to consider the forces on each mass separately. Remember, force is a vector.

Start by considering just the mass on the incline. Draw a free body diagram. What forces are present? Write out an equation for each direction (take the x direction to be along the incline and the y direction to be perpendicular to the incline).

Now do the same for the other mass. This one is much easier, since there are only forces in one direction.

From this you will get a set of equations that you can use to solve for what you need to find.

 

[baba89]

Ft = m2a-m2gCos

sub back to other equation
m1g + (-m2a+m2gCos) = m1a

but i can't get the right answer.

Dear Student,

Thank you for your detailed attempt at solving this problem. It seems like you have a good understanding of the basic equations and concepts related to motion with inclines. However, there are a few mistakes and misunderstandings that are preventing you from getting the correct answer.

Firstly, it is important to note that in this problem, the masses are connected by a string that passes over a frictionless pulley. This means that the tension in the string will be the same for both masses, and can be calculated using the equation Ft = m2a - m2gcosθ, where θ is the angle of the incline. Therefore, your attempt at setting up equations for the tension for each mass separately is not necessary.

Secondly, in your attempt to find the acceleration of m1, you used the equation Fnet = ma, which is correct. However, the forces acting on m1 are not just its weight (mg), but also the tension in the string (Ft). Therefore, the correct equation would be Fnet = m1a = Ft - m1g. Solving for a, we get a = (Ft - m1g)/m1. Now, since we know that the tension in the string is the same for both masses, we can substitute the value of Ft from the previous equation (Ft = m2a - m2gcosθ) and solve for a.

Thirdly, you mentioned that you got an answer of 6.8m/s^2 for the acceleration of m1. This is incorrect, as the acceleration of m1 should be equal to the acceleration of m2 since they are connected by a string. Therefore, the acceleration of both masses should be 0.125m/s^2.

Finally, to find the coefficient of kinetic friction between m2 and the incline, we can use the equation Ff = μkN, where Ff is the force of friction, μk is the coefficient of kinetic friction, and N is the normal force. In this case, the normal force will be equal to the weight of m2, which is m2g. Therefore, the equation becomes Ff = μkm2g. We can