Recent content by bbsgirl10

Mrod = 0.025kg, m = 0.075kg, r1 =0.165m r2 =0.35m, T (String Tension) = 1.21N, Angle Theta = 40 degree. Distance from pivot to center hole (r1) Distance from pivot to brass mass (r2) 1) Equilibrium condition to calculate the horizontal and vertical components of the force exerted by the pivot on...

Theta refers to the angle measured by the sensor minus 90 degree which is 40 degree. I m so lost right now what are the factors involved in Fx and Fy? If the sum is zero do I assume the factors that affect also zero?

:nb)

Tension ?

But why T is not horizontal?

Oh i see then what are the forces acting on Fx and Fy ?

i mean the force acting on Fx Is tension on the string which is why I put r1 cosθ

For horizontal equilibrium: ∑𝐹𝑥 = 0 (Summation of horizontal forces equal zero) T - Fx = 0 Fx = T where Fx= 𝑟1 cos θ For vertical equilibrium: ∑𝐹𝑦 = 0 -m(g) - Mrod (g) + Fy = 0 Fy = mg + Mrod (g) is this correct?

T is a string tension i know the sum of the force will be zero but I don't get it how to calculate fx and fy

τ1= T r1Sinθ and τ2=( mgr2+Mrodgr1) Cos θ

may i know why? because i get around the same value for acceleration and initial velocity based on experiment I got a = 0.200 (for postion/time graph) and a =0.203 (for velocity/time graph). I also get Initial velocity = 0.0711 (for postion/time graph) and Initial velocity = 0.0811 (for...

The graph is just an example to show the type of motion it travels. I just want to know whether the acceleration and initial velocity will be same for both position/time graph and Velocity time graph