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    Connected masses in equilibrium, mass sliding on an incline

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    6. A hanging 2-kg block begins to accelerate a 4-kg block on a horizontal surface. But a rope connected to the wall at 60.8 degrees brings the block to an equilibrium position. What is the tension (to two decimal places) in the angled rope that is required to keep the block from accelerating? The coefficient of friction between the surface and the 4-kg block is 0.25.

    6a. How does the normal force compare to the weight of the block in this problem? Please explain fully.

    7. A 2.0-kg mass is sliding on a surface inclined at 59.5 degrees. The 2.5-kg hanging mass is allowed to drop from rest. It takes 2.1 seconds for the hanging mass to fall through 1.5 meters and hit the ground. What is the coefficient of friction (to three decimal places) between the sliding mass and the incline?

    7a. If the angle of the incline above is decreased, thereby changing the normal force, what would happen to the coefficient of friction? Please explain fully.

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attachments 1 and 2 for the solution and the accompanying diagrams.

    SOLUTION TO PROBLEM No. 6

    The student has not enclosed any diagram. An assumed diagram to described the problem is attached (please see diagram 1).

    Let the tension in the rope connecting the 2 kg mass and 4 kg mass be T2 and that in the rope connecting the 4 kg mass and the wall be T1.

    Tension T1 acting on the 4 kg mass can be resolved into following components :

    Horizontal - T1 sin θ
    Vertical - T1 cos θ where θ = 60.8 degrees

    The net downward force on the 4 kg block = m2g - T1 cos θ. The 4 kg block pushes the horizontal surface with a net force equal to (m2g - T1 cos θ) and therefore the normal reaction R = m2g - T1 cos θ.

    Since frictional force f = μR (where μ is the coefficient of friction), in the present case
    f = μ (m2g - T1 cos θ). ..............(1)

    For equilibrium of 2 kg mass net force on the same must be zero. Thus,

    m1g = T2 ...............(2)

    Also for equilibrium of 4 kg mass :

    T2 - f - T1 sin θ = 0 ....................(3)

    Substituting for f and T2 from (1) and (2) in (3) we get :

    m1g - μ (m2g - T1 cos θ) - T1 sin θ = 0

    or T1 = (μ m2g - m1g)/ (μ cos θ - sin θ)

    Putting μ = 0.25, m1 = 2 kg, m2 = 4 kg, θ = 60.8 degrees, g = 9.8 m/s/s , we get

    T1 = 13.05 N

    ANSWER TO 6a

    The normal reaction force is equal to the force with which the block pushes the surface down. In the present case the block is experiencing an upward force equal to T1 cos θ due to the rope in addition to the weight acting downward. Therefore net downward push of the block on the surface is less than the weight. Thus, the normal reaction in less than the weight.

    SOLUTION TO PROBLEM No. 7

    The student has not enclosed any diagram. An assumed diagram to described the problem is attached (please see diagram 2).

    As the mass m2 takes 2.1 sec to fall by a distance of 1.5 m, starting from rest, applying the equation S = ut + ½ a t2 we get :

    1.5 = 0 + ½ a (2.1)(2.1)

    Or a = 0.68 m/s/s ...................(1)

    From Newtons second law net force acting on mass m2 = F2 = m2 x a = 2.5 x 0.68 = 1.7 N ...................(2)

    Further, F2 is also equal to the difference between the downward weight and upward tension T. Thus, F2 = m2g - T = 1.7 N ..............(3)

    Or 2.5 x 9.8 - T = 1.7

    Or T = 22.8 N .............(4)

    The weight of the 2 kg mass acting vertically downwards can be resolved into two components as follows :

    Parallel to the inclined surface - m1g sinӨ
    Perpendicular to the inclined plane - m1g cosӨ

    Normal reaction on the 2 kg block = R = m1g cosӨ

    Friction force acting on the block = f = µR = µ m1g cosӨ ........(5)

    Net force acting on the 2 kg block parallel to the inclined plane = F1 = T - f - m1g sinӨ

    Or F1 = T - µ m1g cosӨ - m1g sinӨ

    From Newtons second law F1 = m1 x a

    Or T - µ m1g cosӨ - m1g sinӨ = m1 x a ...........(6)

    Putting various values in (6) we get

    22.8 - µ x 2 x 9.8 x cos 59.5O - 2 x 9.8 x sin 59.5O = 2 x 0.68

    Solving we get µ = 0.457

    ANSWER TO 7a

    Coefficient of friction is independent of the angle of incline. It will remain unchanged with change in Ө. However, the friction force acting on the block will change as the normal reaction changes.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:35 pm ad1c9bdddf>
    https://brainmass.com/physics/equilibrium/connected-masses-equilibrium-mass-sliding-incline-56542

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