Answer:
Explanation:
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.
