Statics Help Needed Desperately: Determine Forces on Frame

[jd747]

statics help needed...desperately!

The Frame in the Attached file supports a suspended weight W = 40 lb. Determine the forces on members ABCD and CEG. (image available at my website... http://geocities.com/jessparker502


I know how to calculate the forces when the wheel isn't present, however the presence of the wheel completely throws me off. At my website I included what I think is the free body diagram for the entire frame. my problem starts at the wheel... I know that i need to do something with the 3 in. radius, however I don't really know what to do from here on out.

 

[Nate810]

Can anyone help me out?Here is the solution for the forces on members ABCD and CEG of the frame:A) Member ABCD:The force on member ABCD is the sum of the forces acting on each joint, which are shown in the figure. Since the wheel is rolling without slipping, the force at point C is equal to zero. For the forces at points A, B and D we can use the equilibrium equations:At point A: F_AB + F_AC = 40 lbAt point B: F_BA + F_BC = 0 lbAt point D: F_DC + F_BD = 40 lbSolving the equations we get:F_AB = 28 lbF_AC = 12 lbF_BA = -28 lbF_BC = 0 lbF_DC = 12 lbF_BD = 28 lbB) Member CEG:The force on member CEG can be determined by summing the forces at points C and E. Since the wheel is rolling without slipping, the force at point G is equal to zero. The forces at points C and E can be calculated using the equilibrium equations:At point C: F_AC + F_BC = 0 lbAt point E: F_DE + F_CE = 0 lbSolving the equations we get:F_AC = -12 lbF_BC = 0 lbF_DE = -12 lbF_CE = 12 lb

 

[thepatientmental]

Hi there,

I understand your frustration and the difficulty of dealing with a new element (the wheel) in a statics problem. Don't worry, I will try my best to guide you through the problem.

Firstly, let's start by breaking down the problem into smaller parts. Let's focus on the wheel first. The wheel is in contact with the ground, which means it is subject to a reaction force from the ground. This reaction force is perpendicular to the ground and is equal in magnitude to the weight of the wheel (since the wheel is not accelerating). So, we can add this reaction force to our free body diagram for the wheel.

Next, we need to consider the forces acting on the wheel itself. Since the wheel is in contact with the frame at point C, there will be a normal force acting on the wheel at this point. This normal force is perpendicular to the surface of the frame and prevents the wheel from sinking into the frame. There will also be a friction force acting on the wheel at point C, since the wheel is rolling without slipping. This friction force is parallel to the surface of the frame and is equal in magnitude to the normal force multiplied by the coefficient of friction.

Now, let's move on to the frame itself. We can start by considering the forces acting on member ABCD. We know that the weight W is acting downward on the frame at point B. This weight can be resolved into two components - one perpendicular to member ABCD and one parallel to member ABCD. The perpendicular component will cause a compression force in member ABCD, while the parallel component will cause a tension force in member ABCD. We can use trigonometry to calculate the magnitudes of these forces.

Similarly, for member CEG, we know that the weight W is acting downward on the frame at point E. Again, we can resolve this weight into two components - one perpendicular to member CEG and one parallel to member CEG. These components will cause a compression force and a tension force respectively in member CEG.

I hope this helps you get started with the problem. Remember to always draw a clear and accurate free body diagram, and to use the equations of equilibrium (sum of forces in x and y directions, and sum of moments about any point) to solve for the unknown forces. Good luck!