## Mastering Physics Solutions: Two Masses, a Pulley, and an Inclined Plane

Two Masses, a Pulley, and an Inclined Plane

Part A = (a + gsin(θ) + μgcos(θ)) / (g – a)

Solution Below:

Block 1, of mass m1, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is μ.

Part A

Find the ratio of the masses m1 / m2.

To solve this problem, find the tension with respect to both masses, and then equate, since the tension is the same through the entire rope:

For mass 1:

(m1)g – T = (m1)a
T = (m1)g – (m1)a
T = (m1)(g – a)

For mass 2:

T – (m2)gsin(θ) – (m2)μgcos(θ) = (m2)a
T = (m2)a + (m2)gsin(θ) + (m2)μgcos(θ)
T = (m2)(a + gsin(θ) + μgcos(θ))

Now equate. Note how m1 and m2 were factored in the above equations. This was necessary to isolate m1/m2:

(m1)(g – a) = (m2)(a + gsin(θ) + μgcos(θ))
m1/m2 = (a + gsin(θ) + μgcos(θ)) / (g – a)

(a + gsin(θ) + μgcos(θ)) / (g – a)

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### 12 Responses to Mastering Physics Solutions: Two Masses, a Pulley, and an Inclined Plane

1. Erin says:

(Figure 1) Block 1, of mass m1 = 0.500kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.300, an acceleration of magnitude a = 0.600m/s2 is observed for block 2.

I calculated the mass for block 2 to be 1.56kg. Here is the response I got:

Incorrect;You’ve indicated that the force of friction and the component of the block’s weight parallel to the incline act in different directions.

Any help would be greatly appreciated! thanks!

• Mastering Physics Solutions says:

Yes you are absolutely right! Maybe you were looking at the problem we linked to here. We fixed that problem so it should give the correct answer now.

2. Geoff says:

A 10kg block attached to a 5kg block over an ideal pulley slides down a frictionless surface angled at 40 degrees. What is its acceleration?

• Mastering Physics Solutions says:

Well, it depends on the angle of the ramp, for example.

3. Courtney says:

Prompt: Block 1, of mass m1 = 0.650kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.350, an acceleration of magnitude a = 0.250m/s2 is observed for block 2.

• Mastering Physics Solutions says:

Check out the solution to your problem here.

4. Marcus says:

First of all I would like to thank you guys for this AWESOME website!!
Second: this problem was a little different in my homework, they might have changed it since last year. So here goes the new prompt and the question:

Prompt: Block 1, of mass m1 = 0.650kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.200, an acceleration of magnitude a = 0.450m/s2 is observed for block 2.

Part. A: Find the mass of block 2, m2.

Thank You!!

• Mastering Physics Solutions says:

For your numbers, the answer should be 1.67 kg. We just posted a detailed solution here. Hope this helps!

• Samantha says:

Prompt: Block 1, of mass m1 = 0.650kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.200, an acceleration of magnitude a = 0.200m/s2 is observed for block 2.

Part. A: Find the mass of block 2, m2.

• Mastering Physics Solutions says:

We posted a solution to this problem here. The acceleration in your version of the problem is different, but you can follow the same method we used and get the right answer. Off the cuff, it should be about 1.84 kg, but double check.

5. Angie says:

OMG! Thank you soooooo much! this was really helpful!

• Mastering Physics Solutions says:

You’re very welcome, glad to help!