Recent content by jojosg

I'm not confident in my acceleration Diagram because this question should have coriolis acceleration but I don't know how to draw it in acceleration diagram. Help would be appreciated. $$V_D = 4.1 \, \text{cm} \times (2 \, \text{m/s}) / 1 \, \text{cm} = 8.2 \, \text{m/s}$$ $$\omega_{BC} =...

Am I doing this correctly? Given variables: ##R_b## = base circle radius ##R_f## = roller radius ##e## = offset (positive as shown in figure) ##\Delta R## = follower displacement from home position ##\frac{\nu}{\omega_{\rm{cam}}} = \frac{ds}{d\theta}## = slope of displacement diagram...

Textbook This is the textbook I'm using for this course. The original position I refer to is based of the diagram provided.

For the 134 degrees I drew the case where link AB and link BC are coincident and fold together. The I measured the angle between AB original position to the folded position which gives me 134 degrees. Maximum pressure angle occurs when the transmission angle is minimal, typically at extreme...

I'm also not not sure about 134 degrees. I'm not sure how to measure pressure angle

Did I do mine correctly?

Just unsure if I did it correctly. I identified G as an idle Degree of Freedom. Should I exclude it from DOF calculation? But is I do that The Current DOF would be more than 2 which is definitely the wrong answer.

ok Thank you very much!

Hi, just need somebody to check my homework solution. Not entirely sure of my answer. My teacher used Mechanics of Materials by RC Hibbler, but I could not find an example question to solve this homework question. Equilibrium at pin B Take moments about B: ## F_1 \cdot 40 = F_2 \cdot...

I used basic trigo and drew a triangle with adjacent and opposite sides both a to get 45 degrees

Ok I think I got it. ##\sum F_x = 0 : \quad A_x - B_x = -F \tag{1}## ##\sum F_y = 0 : \quad A_y + B_y = 0 \tag{2}## ##\sum M_A = 0 : \quad B_y(2a) - F(2a) = 0 \tag{3}## ##\sum M_E = 0:## ##-\frac{\sqrt{2}}{2}F_{DG}(a) - \frac{\sqrt{2}}{2}F_{DG}(a) - F(a) = 0 \tag{1}## ##-\sqrt{2}F_{DG} - F = 0...

Been trying for a While. All I know after these is to take moment about O in the sectioned FBD but I keep getting redundant equation that do not let me find internal force Toa. 1. Global Equilibrium Sum of moments about point B: ##\sum M_B = 0## ##-F(2a) + A_y(2a) = 0## ##A_y = F## Sum of...

\sum M_0 = 0 -F(2a) + A_y(2a) = 0 A_y = F \sum F_x = 0 F + A_x = O_x \sum F_y = 0 O_y + A_y = 2F \sum M_A = 0 -F(2a) + 2F(2a) - O_y(2a) = 0 F = O_y F = O_x - A_x \sum F_y = 0 -2F - \frac{F_{OH}}{\sqrt{2}} = 0 O_x = 2F A_x = F Summary: \boxed{A_y = F} \boxed{A_x = F} \boxed{O_x = 2F}...

Member BE: \sum F_x = 0 \quad \Rightarrow \quad \frac{F_G}{\sqrt{2}} = F + E_x \sum F_y = 0 \quad \Rightarrow \quad E_y = \frac{F_G}{\sqrt{2}} \Rightarrow F_G = \sqrt{2}E_y From equilibrium: E_y = F + E_x Member BC: \sum F_x = 0 \quad \Rightarrow \quad B_x = E_x + \frac{F_C}{\sqrt{2}}...