1. The problem statement, all variables and given/known data a ladder of length L and mass m is leaning against a wall as shown in the figure below. assuming the wall and the floor are frictionless, the ladder will slide down the floor and until the left end loses contact with the wall. Before the ladder loses contact with the wall there is one degree of freedom θ. 1.Express the kinetic energy interms of θ and dθ/dt. 2.find the virtual work and the generalized force. 2. Relevant equations Note: the bold symbols are related to vectors. T=1/2 ×i=1∑i=M (mi×dsi/dt⋅dsi/dt)=T(q1,.......,qn,dq1/dt,.....,dqn/dt).⇒⇒⇒formula of kinetic energy. (1) M is all parts of the system. δw=k∑×(i∑Fi*∂ri/∂qk)δqk.⇒⇒⇒formula of virtual work. (2) k is the number of degrees of freedom,and qk is a degree of freedom i∑Fi*∂ri/∂qk=Fk(generalized force). (3) Fk=d/dt(∂T/∂(dqk/dt))-∂T/∂qk.(4) 3. The attempt at a solution I used Cartesian coordinates, the part related to the ground is the x axis and how the ladder is high from the ground is the y axis. Let the origin be the point of contact of the ladder with the ground and "s" is how each infinitesimal part is far from the origin. 1.Let "h" be the distance between the point of contact of the ladder with the ground and the wall, let "y" the distance between the point of contact of the ladder with the wall and the ground. The vector Si=hii+yij. dsi/dt=dhi/dti+dyi/dtj. h=cosθs. y=sinθs. si.si=(dθ/dt)2 *(sinθ)2*s2 +(dθ/dt)2*(cosθ)2*s2 (dsi/dt)2=s2 *(dθ/dt)2 T=0∫L (dT)=0∫L (1/2*dm*(ds/dt)2)=0.5*0∫L(M/L*ds*s2*(dθ/dt)2)=0.5*(dθ/dt)2*M/L*0∫Ls2*ds=1/6*(dθ/dt)2*ML2 2. I calculated here the generalized force. ∂ri/∂θ=∂h/∂θi+∂y/∂θj=-(dθ/dt)sinθsi+(dθ/dt)cosθsj Using contious form of equation (3) we get : Fk=(intergrate from zero to L) ∫dmgj.(-(dθ/dt)*sinθ*si +dθ/dt*cosθ*sj) I get finally Fk=Mg(dθ/dt)*cosθ*(L/2). when I use equation (4) to get the generalized force I get different answer, so is there something wrong in my solution?