Recent content by ptguard1

Wow, I just realized how much more sense that makes. Thank you.

∫∫9(x + y) e^(x2 − y2) dA, where R is the rectangle enclosed by the lines x−y=0, x−y=10, x+y=0, and x+y=5 Relevant Equations: The Jacobian: ∂(x,y)/∂(u,v) The attempt at a solution: I began by making u=x+y and v=x^2-y^2 So, u=0 and u=5, but I don't know what to do with the x-y line...

Wonderful! I performed the integration and got 10ln(√3) and this was correct.

∫∫10xy(dA), where R is the region in the first quadrant bounded by the lines y=x/2 and y=2x/3 and by the hyperbolas xy=1/2 and xy=3/2 The transformations given in the problem (these cannot be altered): x=u/v and y=v Relevant equations: The Jacobian - ∂(x,y)/∂(u,v) The attempt at a solution...

I just discovered that I actually didn't need to find the forces at each pin in order to solve the overall problem, but thank you for the information.

Homework Statement View image below Homework Equations The Attempt at a Solution I don't understand where the normal and shear forces are applied. Are there one of each on both inclined planes or is there just one normal force and one shear force? If you could provide a drawing I...

I recalculated my solutions and found my mistake. F at BC = 165 kN (up and to the right), A_x = 142.89 kN to the right, A_y = 82.5 kN up. The issue I'm having is solving for the shear stress in the pins. I don't understand why all of the pins would have the same result. For the pin at A, I...

Homework Statement View image for all details Homework Equations tao = V/A The Attempt at a Solution I went ahead and solved the force at BC and the x and y components at A, but I feel as though my solutions are incorrect. At BC I have F = 825 kN (up and to the left), A_x = 714.5...

What do you mean by an axial force?

I went ahead and made a rough sketch on paint

Homework Statement In a simple truss where P = 8 kip, find the unknown forces D_x, D_y, C_x, and C_y then determine the average normal stress in each member. Description of truss: I cannot copy an image due to the source it is coming from, so here is my attempt at a description A triangle...

I figured out that the force is equal to the 58.8 N of force that the sack had on the earth. My logic was that an object can be lifted at different speeds with the same force, is this true or only partially true?

A sack of flour of mass 6.00 kg is lifted vertically at a constant speed of 4.00 m/s through a height of 18.0 m How great a force is required? F=ma I found the force that the sack has on the Earth (58.8N), but don't understand how I am suppose to find what force is being applied to...