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30 July, 22:03

The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both of which lie in a horizontal plane. For the position where xA = 16 in., the velocity of A is vA = 3 ft/sec to the right. Determine the acceleration of each slider and the force in the bar at this instant.

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  1. 30 July, 23:38
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    Given:

    - The Length of the rigid bar L = 20 in

    - The position of slider a, x_a = 16 in

    - The position of slider b, y_b

    - The velocity of slider a, v_a = 3 ft / s

    - The velocity of slider b, v_b

    - The acceleration of slider a, a_a

    - The acceleration of slider b, a_b

    Find:

    -Determine the acceleration of each slider and the force in the bar at this instant.

    Solution:

    - The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

    L^2 = x_a^2 + y_b^2 ... 1

    y_b = sqrt (20^2 - 16^2)

    y_b = 12

    - The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

    0 = 2*x_a*v_a + 2*y_b*v_b

    0 = x_a*v_a + y_b*v_b ... 2

    0 = 16*36 + 12*v_b

    v_b = - 48 in / s = - 4 ft/s

    - Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

    0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

    0 = 9 + 4*a_a/3 + 16 + a_b

    4*a_a/3 + a_b = - 25

    4*a_a + 3*a_b = - 75 ... 3

    - Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

    F_x = m_a*a_a

    P - R_r*16/20 = m_a*a_a

    - For Slider B, Apply Newton's second law of motion in y direction:

    F_y = m_b*a_b

    - R_r*12/20 = m_b*a_b

    - Combine the two dynamic equations:

    P - 4*m_b*a_b / 3 = m_a*a_a

    3P = 3*m_a*a_a + 4*m_b*a_b ... 4

    - Where, P = Is the force acting on slider A

    P, m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.
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