To determine The angular acceleration of the link AB . Explanation Draw the schematic diagram of the link AB with instantaneous center as shown in Figure (1). Write the expression for the velocity of point A . v A = O A ¯ O E ¯ v b (I) Here, distance between the points O and A is O A ¯ , distance between the points O and E is O E ¯ and speed of the band is v b . Write the expression for the angular velocity of the bar AB . ω A B = v A A C ¯ (II) Here, angular velocity of the link AB is ω A B . Draw the schematic diagram of the link DB with as shown in Figure (2). Write the expression for the angular velocity of the bar BD . ω B D = v B B D ¯ (III) Here, angular velocity of the link BD is ω B D and distance between the points B and D is B D ¯ . Write the expression for the acceleration of point B in vector form. a B = a A + a B / A ( a B ) n + ( a B ) t = a A + a B / A (IV) Here, acceleration of point B in vector form is a B , normal component of acceleration of point B in vector form is ( a B ) n , tangential component of acceleration of point B in vector form is ( a B ) t , acceleration of point A in vector form is a A and acceleration of point B with respect to point A in vector form is a B / A . Write the expression for the acceleration of point B with respect to point A in vector form. a B / A = α × r B / A − ω A B 2 r B / A (V) Here, angular acceleration of the link AB in vector form is α and position vector of point B with respect to point A is r B / A . Conclusion: Substitute 0.3 m for O A ¯ , 0.4 m for O E ¯ and 2 m / s for v b in Equation (I). v A = 0.3 m 0.4 m ( 2 m / s ) = 1.5 m / s Since the instantaneous center is at infinity for the link AB , the velocity of point B will be same as the velocity of point A . v B = v A = 1.5 m / s Substitute 1.5 m / s for v A and ∞ for A B ¯ in Equation (II). ω A B = 1.5 m / s ∞ = 0 rad / s Substitute 1.5 m / s for v B and 0.18 m for B D ¯ in Equation (III). ω B D = 1.5 m / s 0.18 m = 8.33 rad / s Calculate the position vector of point B with respect to point A . r B / A = A B ¯ ( − cos 15 ° i + sin 15 ° j ) Here, length of the link AB is A B ¯ and unit vectors along x and y directions are i and j respectively. Substitute 0.36 m for A B ¯ . r B / A = 0.36 m ( − cos 15 ° i + sin 15 ° j ) = − 0.3477 i + 0.09317 j m Write the expression for the angular acceleration of the link AB in vector form. α = α A B k Here, angular acceleration of the link AB is α A B and unit vector along z direction is k . Write the expression for the upward acceleration of point A in vector form. a A = v A 2 O A ¯ j Substitute 0

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Chapter 5.6, Problem 148P

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The angular acceleration of the link AB.

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Chapter 5.6, Problem 147P

Chapter 5.6, Problem 149P

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