he uniform 110-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position withθ= 0. Determine the init

Question

he uniform 110-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position withθ= 0. Determine the initial angular accelerationα of the beam and the magnitudeFAof the force supported by the pin at A due to the application of the force P = 350 N on the attached cab

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Philomena 2 months 2021-07-31T07:08:39+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-07-31T07:10:05+00:00

    Answer:

    The initial angular acceleration of the beam is {\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}

    The magnitude of the force at A is 832.56N

    Explanation:

    Here, m is the mass of the beam and l is the length of the beam.

    I =\frac{1}{3} ml

    I=\frac{1}{3} \times110\times4^2\\I=586.67kgm^2

    Take the moment about point A by applying moment equilibrium equation.

    \sum M_A =I \alpha

    P \sin45^0 \times 3 =I \alpha

    Here, P is the force applied to the attached cable and \alpha is the angular acceleration.

    Substitute 350 for P and 586.67kg.m² for I

    350 \sin 45^0 \times3=568.67 \alpha

    \alpha =1.3056rad/s^2

    The initial angular acceleration of the beam is {\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}

    Find the acceleration along x direction

    a_x = r \alpha

    Here, r is the distance from center of mass of the beam to the pin joint A.

    Substitute 2 m for r and 1.3056rad/s² for \alpha

    a_x = 2\times 1.3056 = 2.6112m/s^2

    Find the acceleration along the y direction.

    a_y = r \omega ^2

    Here, ω is angular velocity.

    Since beam is initially at rest,ω=0

    Substitute 0 for ω

    a_y = 0

    Apply force equilibrium equation along the horizontal direction.

    \sum F_x =ma_x\\A_H+P \sin45^0=ma_x

    A_H + 350 \sin45^0=110\times2.6112\\\\A_H=39.75N

    Apply force equilibrium equation along the vertical direction.

    \sum F_y =ma_y\\A_V-P \cos45^0-mg=ma_y

    A_v +350 \cos45^0-110\times9.81=0\\A_V = 831.61\\

    Calculate the resultant force,

    F_A=\sqrt{A_H^2+A_V^2} \\\\F_A=\sqrt{39.75^2+691.61^2} \\\\= 832.56N

    The magnitude of the force at A is 832.56N

    0
    2021-07-31T07:10:36+00:00

    Answer:

    a) Initial angular acceleration of the beam = 1.27 rad/s²

    b) F_{A} = 851.11 N

    Explanation:

    tan \theta = \frac{opposite}{Hypothenuse} \\tan \theta = \frac{3}{3} = 1\\\theta = tan^{-1} 1 = 45^{0}

    Force applied to the attached cable, P = 350 N

    Mass of the beam, m = 110-kg

    Mass moment of the inertia of the beam about point A = I_{A}

    Using the parallel axis theorem

    I_{A}  = I_{G} + m(\frac{l}{2} )^{2} \\I_{G} = \frac{ml^{2} }{12} \\I_{A}  = \frac{ml^{2} }{12}  + \frac{ml^{2} }{4} \\I_{A}  = \frac{ml^{2} }{3}

    Moment = Force * Perpendicular distance

    \sum m_{A} = I_{A} \alpha\\

    From the free body diagram drawn

    \sum m_{A}  = 3 Psin \theta\\ 3 Psin \theta = \frac{ml^{2} \alpha }{3}

    P = 350 N, l = 3+ 1 = 4 m, θ = 45°

    Substitute these values into the equation above

    3 * 350 * sin 45 = \frac{110 * 4^{2}* \alpha }{3} \\\alpha = 1.27 rad/s^{2}

    Linear acceleration along the x direction is given by the formula

    a_{x} = r \alpha

    r = 2 m

    a_{x} = 2 * 1.27\\a_{x} = 2.54 m/s^{2}

    the linear acceleration along the y-direction is given by the formula

    a_{y} = r w^{2}

    Since the beam is initially at rest, w = 0

    a_{y} = 0 m/s^{2}

    General equation of motion along x – direction

    F_{x} + Psin \theta = ma_{x}

    F_{x} + 350sin45 = 110 * 2.54\\F_{x} = 31.913 N

    General equation of motion along y – direction

    F_{x} + Pcos \theta - mg = ma_{y}

    F_{y} + 350cos45 - 110*9.8 = m* 0\\F_{y}  = 830.513 N

    Magnitude F_{A} of the force supported by the pin at A

    F_{A}  = \sqrt{F_{x} ^{2} + F_{y} ^{2} } \\F_{A}  = \sqrt{31.913 ^{2} + 850.513 ^{2} } \\F_{A}  = 851.11 N

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