korosh23
  • korosh23
Physics 12 Rotational Equilibrium Question! Please explain how you get the answer. Wait for the attachment.
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
korosh23
  • korosh23
1 Attachment
korosh23
  • korosh23
|dw:1447370230465:dw|
anonymous
  • anonymous
You need to look at the force components perpendicular to the beam, and balance the moments around the hinge

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anonymous
  • anonymous
|dw:1447385639538:dw|
anonymous
  • anonymous
We have three force components to work with, and one moment. The first step is to find equations that describe the components of force that are perpendicular to the beam, relative to the forces Fg, Fm and Ft For Fg, the force component perpendicular to the beam will be Fg(cos(50)). Where Fg is the force due to the weight of the beam (350 kg) which is approximately 3500 N, given g = 10 m/s^2 For Fm, the force component perpendicular to the beam will be Fm(cos(50)) And finally for the tension force we have Ft(cos(40)) We can now use these forces to balance the moments around the hinge, where we need M1 to be zero As torque is a function of both force and distance, we must multiply our perpendicular force components by the distance the force is acting from the hinge. For the weight of the beam, the force acts through the center of mass, which will be at the center of the beam, so 2.1 m For the tension force the distance will be 3 m and for the Mass, the distance will be 4.2 m We can now write an equation to balance the moments around the hinge. As Fm and Fg are acing on an opposite side of the beam to Ft, Ft in our equation will be negative. In reality it doesn't matter if you take Ft as negative, or Fm and Fg as negative, as long as they are opposing. This gives us the equation: Fg(cos(50))(2.1) + Fm(cos(50))(4.2) - Ft(cos(40))(3) = M1 Where: Fg = 3500 N Ft = 1.3x10^4 N M1 = 0 N m We can now plug these values into the equation, and solve for Fm. 3500(cos(50))(2.1) + Fm(cos(50))(4.2) - (1.3x10^4)((cos(40))(3) = 0 4724.5 N + 2.7Fm N - 29875.7 N = 0 2.7*Fm N = 25151.2 Fm = 9315 N This gives us a force in Newtons, but the questions asks for a Mass, Kg, so we just divide our answer by g = 10 to give our answer in Kg 931.3 Kg Hope this helps
korosh23
  • korosh23
are you physics grade 12 student ?
anonymous
  • anonymous
I don't know what that means
korosh23
  • korosh23
Are you university student or physics student or professsor, could you please answer my question. It is imp
anonymous
  • anonymous
University student
korosh23
  • korosh23
Ok do you know about the position of folcrum and how to find the force of wall in the y direction, it is a different question, if yes can you helo me with something it is very confusing for me.
korosh23
  • korosh23
?
korosh23
  • korosh23
I am not sure if the internet connection is slow, but do you know about it?
anonymous
  • anonymous
Yeah my internet connection is terrible in New Zealand. To determine the force of the wall in the Y direction, take a sum of the forces in the Y direction of the entire system. The only forces in the Y direction in the system are Fg and Fm, so the force in the Y direction in the wall, Fwy = Fg + Fm
anonymous
  • anonymous
Yeah my internet connection is terrible in New Zealand. To determine the force of the wall in the Y direction, take a sum of the forces in the Y direction of the entire system. The only forces in the Y direction in the system are Fg and Fm, so the force in the Y direction in the wall, Fwy = Fg + Fm
korosh23
  • korosh23
Yeah I understand lol but what if we have do not have the value of the forces, and we are asked to find torques, do I have to find the net torque and then the Force of wall in y-direction?
anonymous
  • anonymous
I'm not sure I understand your question, sorry. If you don't have any force values, then you can not determine torque, as torque is a function of both force and distance.
korosh23
  • korosh23
Ok @PeterOrsome thank you for your help and time to answer my question. If I have further questions, I will tag you in my question. you are really patient with that internet connection hahaha joking

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