Answer:
49 N
Explanation:
The diagram of the bar is obtained online and attached to this solution.
The free body diagram is also attached.
Since the weight of the bar acts at the middle of the bar, the torque due to the weight of the bar is given by
τ = mgx
where x = (L/2) cos 35° = 0.45 × cos 35° = 0.3686 m
τ = (7)(9.8)(0.3686) = 25.29 Nm
The force acting on pin A = torque ÷ (length × sin 35°) = 25.29 ÷ (0.9 × sin 35°)
= 25.29 ÷ 0.5162 = 48.99 N = 49 N
Hope this Helps!!!
Answer:
The force supported by the pin at A is 69.081 N
Explanation:
The diagram is in the figure attach. The angular acceleration using the moment expression is:
[tex]-mg(\frac{Lcos\theta }{2} )=I\alpha \\\alpha =\frac{-3g}{2L} cos\theta[/tex]
Where
L = length of the bar = 900 mm = 0.9 m
[tex]\alpha =\frac{-3*9.8cos35}{2*0.9} =-13.38rad/s^{2}[/tex]
The acceleration in point G is equal to:
[tex]a_{G} =a_{A} +\alpha kr_{G/A} -w^{2} r_{G/A}[/tex]
Where
aA = acceleration at A = 0
w = angular velocity of the bar = 3 rad/s
rG/A = position vector of G respect to A = [tex]\frac{L}{2} cos\theta i-\frac{L}{2} cos\theta j[/tex]
[tex]a_{G} =(\frac{L}{2}\alpha sin\theta -\frac{w^{2}Lcos\theta }{2} )i+(\frac{L}{2}\alpha cos\theta +\frac{w^{2}Lsin\theta }{2} )j=(\frac{0.9*(-13.38)*sin35}{2} -(\frac{3^{2}*0.9*cos35 }{2} )i+(\frac{0.9*(-13.38)*cos35}{2} +\frac{3^{2} *0.9*sin35)}{2} )j=-6.76i-2.61jm/s^{2}[/tex]
The force at A in x is equal to:
[tex]A_{x} =ma_{G} =7*(-6.76)=-47.32N[/tex]
The force at A in y is:
[tex]A_{y} =ma_{G} +mg=(7*(-2.61))+(7*9.8)=50.33N[/tex]
The magnitude of force A is equal to:
[tex]A=\sqrt{A_{x}^{2}+A_{y}^{2} } =\sqrt{(-47.32^{2})+50.33^{2} } =69.081N[/tex]
