Explore BrainMass

Explore BrainMass

    Connected masses in equilibrium, mass sliding on an incline

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

    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.

    © BrainMass Inc. brainmass.com March 4, 2021, 6:41 pm ad1c9bdddf
    https://brainmass.com/physics/equilibrium/connected-masses-equilibrium-mass-sliding-incline-56542

    Attachments

    Solution Preview

    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 ...

    Solution Summary

    Problem 6 : Condition for equilibrium of various forces acting on the block is considered (diagram included).

    Problem 7 : Forces acting on the connected masses - one hanging and the other on the inclined plane are considered (diagram included).

    $2.49

    ADVERTISEMENT