Question

In the figure below the two blocks are connected by a string of negligible mass passing over a frictionless pulley. m1 = 10.0 kg and m2 = 4.50 kg and the angle of the incline is θ = 44.0°. Assume that the incline is smooth. (Assume the +x direction is down the incline of the plane.)

(a) With what acceleration does the mass m2 move on the incline surface? Indicate the direction with the sign of your answer.

(b) What is the tension in the string?

(c) For what value of m1 will the system be in equilibrium?

Answer 1

Solution: The mass of m1 = 10.0 kg

The mass of m₂ = 4.50 kg

Angle of inclination ? = 44.0^o

The free body diagram is as shown below,

Since incline is smooth, there are no frictional forces acting on it. The tension T in the string is uniform along its length.

Part (a)

We have shown different forces acting on the masses. The component of weight of m₂ down the incline causes it to accelerate with an acceleration a.

We use Newton’s second law of motion;

m₂*g*sin? – T = m₂*a ---------------------------------------------- (1)

T = m₁*a                     -------------------------------------------------(2)

It means total force of m₂ (component of weight along incline minus tension) causes it to accelerate. The same tension causes the mass m₁ to accelerate.

Solving above equation (1) and (2) for a, we get,

m₂*g*sin? – m₁*a = m₂*a

m₂*g*sin? = (m₂+m₁)*a

a = m₂*g*sin? / (m₂+m₁)

a = 4.50kg*(9.81m/s²)*sin(44^o)/(4.50+10.0)kg

a = 2.1149 m/s²        or

a = + 2.12 m/s²   along the incline

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Part (b)

Solving equation (2) for T from the obtained a,

T = m₁*a

T = 10.0 kg*2.1149 m/s²

T = 21.149 N     or

T = 21.15 N

Thus tension in the string is T = 21.15 N

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Part (c)

Observe equation (1). If the tension T equals to the component of weight of m2 along the incline, then net force is zero on the left side of equation; hence zero acceleration. Thus we want

m₂*g*sin? – T = m₂*a

0 = m₂*a      that is

m₂*g*sin? – T = 0

Now we have T = m₁*a, thus above equation becomes,

m₂*g*sin? – T = 0

m₂*g*sin? = m₁*a

m₁ = m₂*g*sin? / a

since we want a = 0 m/s² that is equilibrium condition,

m₁ = m₂*g*sin? / 0

m₁ = ? kg.

Thus system will be in equilibrium only when m₁ is infinitely large.