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Question

Two blocks on two ramps are connected by a string passing over two massive, frictionless pulleys. The block on the left has a mass of 190 g and moves up a ramp that is inclined at an angle of 29∘ from the horizontal; the block on the right has a mass of 370 g and moves down a ramp that is inclined at an angle of 51∘ from the horizontal. Both of the blocks experience kinetic friction, with the same coefficient of 0.31. The pulley on the left is a 95 g, 2 cm radius thin ring; the pulley on the right is a 110 g, 4 cm radius thin solid disk.

a) Determine the magnitude of the acceleration of the system, in m/s^2

b) Determine the angular acceleration of the ring-shaped pulley, in rev/s^2

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Diagram: https:// usb06ca.theexpertta.com/images/zt2btrp1.jqo.png

Answer 1

Let

M₁ = 190g be the mass of the block on the left,

M2 = 370g be the mass of the block on the right,

θ₁ = 29° be the angle of the ramp on the left,

θ2 = 51° be the angle of the ramp on the right,

μ = 0.31 be the coefficient of friction,

m₁ = 95g be the mass of the left pully,

r₁ = 2cm be the radius of the left pully,

I₁ = m₁r₁² be the moment of inertia of the left pully,

m2 = 110g be the mass of the right pully,

r2 = 4cm be the radius of the right pully,

I2 = m₁r₁²/2 be the moment of inertia of the right pully,

g = 9.81 m/s² be gravitational acceleration,

a (to be computed) be the acceleration of the system.

Please try to draw the following forces in the diagram.

Let

G₁ = M₁g be the vertical weight of the block on the left,

G2 = M2g be the vertical weight of the block on the right,

G₁cos(θ₁) is the component of G₁ perpendicular to the platform,

G₁sin(θ₁) is the component of G₁ parallel to the platform,

G2cos(θ2) is the component of G2 perpendicular to the platform,

G2sin(θ2) is the component of G2 parallel to the platform,

F₁ = G₁cos(θ₁)μ is the force of friction acting on the left block;

it is acting against movement, that is, parallel to the platform directed down,

F2 = G2cos(θ2)μ is the force of friction acting on the right block;

it is acting against movement, that is, parallel to the platform directed up,

T₁ (unknown) is the tension on the string between the left block and the left pully.

There are two forces of magnitude T₁, both along the string.

One force of magnitude T₁ is acting on the left block pulling it up.

The other force of magnitude T₁ is acting on the left pully downward.

T2 (unknown) is the tension on the string between the right block and the right pully.

There are two forces of magnitude T2, both along the string.

One force of magnitude T2 is acting on the right block pulling it up.

The other force of magnitude T2 is acting on the right pully downward.

T (unknown) is the tension on the string between the two pullies.

There are two forces of magnitude T, both along the string.

One force of magnitude T is acting on the left pully directed up.

The other force of magnitude T is acting on the right pully directed down.

On the assumption that the string does not stretch,

both blocks experience the acceleration a,

and so does the string.

Therefore

α₁ = a/r₁ is the angular acceleration of the left pully,

α2 = a/r2 is the angular acceleration of the right pully.

Notice that in the above there are 4 unknown quantities: a, T1, T2, T.

We know write 4 equations that will allow us to solve for those 4 quantities.

The equations express Newton's second law for those 4 objects involved.

The block on the left:

M1a = T1 - G1sin(θ1) - F1

where the right hand side is the net force pulling the block up.

The block on the right:

M2a = -T2 + G2sin(θ2) - G2cos(θ1)μ

where the right hand side is the net force pulling the block down.

The pully on the left:

I₁α₁ = r₁(T - T₁)

where the right hand side is the net torque turning the pully counterclockwise

The pully on the right:

I2α2 = r2(T2 - T)

where the right hand side is the net torque turning the pully clockwise

Here are the four equations after substituting given quantities

M₁a = T₁ - M₁gsin(θ₁) - M₁gcos(θ₁)μ

M2a = -T2 + M2gsin(θ2) - M2gcos(θ2)μ

m₁r₁²a/r₁ = r₁(T - T₁)

m2r2²a/2r2 = r2(T2 - T)

After simplification

T₁ = M₁(a + gsin(θ₁) + gcos(θ₁)μ)

T2 = M2(-a + gsin(θ2) - gcos(θ2)μ)

T = T₁ + m₁a

T2 = T + m2a/2

After substituting T₁, T2, T

M₁(a + gsin(θ₁) - gcos(θ₁)μ) + m₁a + m2a/2 = M2(-a + gsin(θ2) - gcos(θ2)μ)

a = g(-M₁sin(θ₁) - M₁cos(θ₁)μ + M2sin(θ2) - M2cos(θ2)μ)/(M₁ + M2 + m₁ + m2/2)

The above formula has the following physical interpretation.

As expected, the acceleration a is a ration between forces causing acceleration, and

masses impeding acceleration.

"-M₁sin(θ₁)" is the negative contribution of gravity of the left box.

"-M₁cos(θ₁)μ" is the negative contribution of friction acting on the left box.

"M2sin(θ2)" is the positive contribution of gravity of the right box.

"-M2cos(θ2)μ" is the negative contribution of friction acting on the right box.

"M₁ + M2 + m₁ + m2/2" are all the masses impeding acceleration.

The right pully contributes only half of is mass because the mass is distributed throughout the disk.

(a) You can substitute actual numbers the get the acceleration a, which is approximately 1m/s².

(b) The angular acceleration of the left pully is

α₁ = a/r₁ = (1m/s²)/(2cm) = 50 1/s² = (50/2π) rev/s² ≈ 8 rev/s²