Answer:
The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]
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.
[tex]I =\frac{1}{3} ml[/tex]
[tex]I=\frac{1}{3} \times110\times4^2\\I=586.67kgm^2[/tex]
Take the moment about point A by applying moment equilibrium equation.
[tex]\sum M_A =I \alpha[/tex]
[tex]P \sin45^0 \times 3 =I \alpha[/tex]
Here, P is the force applied to the attached cable and [tex]\alpha[/tex] is the angular acceleration.
Substitute 350 for P and 586.67kg.m² for I
[tex]350 \sin 45^0 \times3=568.67 \alpha[/tex]
[tex]\alpha =1.3056rad/s^2[/tex]
The initial angular acceleration of the beam is [tex]{\bf{1}}{\bf{.3056}}\,\frac{{{\bf{rad}}}}{{{{\bf{s}}^{\bf{2}}}}}[/tex]
Find the acceleration along x direction
[tex]a_x = r \alpha[/tex]
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 [tex]\alpha[/tex]
[tex]a_x = 2\times 1.3056 = 2.6112m/s^2[/tex]
Find the acceleration along the y direction.
[tex]a_y = r \omega ^2[/tex]
Here, ω is angular velocity.
Since beam is initially at rest,ω=0
Substitute 0 for ω
[tex]a_y = 0[/tex]
Apply force equilibrium equation along the horizontal direction.
[tex]\sum F_x =ma_x\\A_H+P \sin45^0=ma_x[/tex]
[tex]A_H + 350 \sin45^0=110\times2.6112\\\\A_H=39.75N[/tex]
Apply force equilibrium equation along the vertical direction.
[tex]\sum F_y =ma_y\\A_V-P \cos45^0-mg=ma_y[/tex]
[tex]A_v +350 \cos45^0-110\times9.81=0\\A_V = 831.61\\[/tex]
Calculate the resultant force,
[tex]F_A=\sqrt{A_H^2+A_V^2} \\\\F_A=\sqrt{39.75^2+691.61^2} \\\\= 832.56N[/tex]
The magnitude of the force at A is 832.56N
Answer:
a) Initial angular acceleration of the beam = 1.27 rad/s²
b) [tex]F_{A} = 851.11 N[/tex]
Explanation:
[tex]tan \theta = \frac{opposite}{Hypothenuse} \\tan \theta = \frac{3}{3} = 1\\\theta = tan^{-1} 1 = 45^{0}[/tex]
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 = [tex]I_{A}[/tex]
Using the parallel axis theorem
[tex]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}[/tex]
Moment = Force * Perpendicular distance
[tex]\sum m_{A} = I_{A} \alpha\\[/tex]
From the free body diagram drawn
[tex]\sum m_{A} = 3 Psin \theta\\ 3 Psin \theta = \frac{ml^{2} \alpha }{3}[/tex]
P = 350 N, l = 3+ 1 = 4 m, θ = 45°
Substitute these values into the equation above
[tex]3 * 350 * sin 45 = \frac{110 * 4^{2}* \alpha }{3} \\\alpha = 1.27 rad/s^{2}[/tex]
Linear acceleration along the x direction is given by the formula
[tex]a_{x} = r \alpha[/tex]
r = 2 m
[tex]a_{x} = 2 * 1.27\\a_{x} = 2.54 m/s^{2}[/tex]
the linear acceleration along the y-direction is given by the formula
[tex]a_{y} = r w^{2}[/tex]
Since the beam is initially at rest, w = 0
[tex]a_{y} = 0 m/s^{2}[/tex]
General equation of motion along x - direction
[tex]F_{x} + Psin \theta = ma_{x}[/tex]
[tex]F_{x} + 350sin45 = 110 * 2.54\\F_{x} = 31.913 N[/tex]
General equation of motion along y - direction
[tex]F_{x} + Pcos \theta - mg = ma_{y}[/tex]
[tex]F_{y} + 350cos45 - 110*9.8 = m* 0\\F_{y} = 830.513 N[/tex]
Magnitude [tex]F_{A}[/tex] of the force supported by the pin at A
[tex]F_{A} = \sqrt{F_{x} ^{2} + F_{y} ^{2} } \\F_{A} = \sqrt{31.913 ^{2} + 850.513 ^{2} } \\F_{A} = 851.11 N[/tex]
