Recent content by Versaiteis

Oh I see, the two systems are directly related so I can simply say L = L_{1} + L_{2} L_{1} = 2.8 * (-0.5 * -3.00) L_{2} = 4.0 * ( 0 - (-0.1 * 4.55)) Sure enough L = 6.02\widehat{k} kg*m^2/s Thank you for your help Andrew

Homework Statement Homework Equations L = m( r x v ) The Attempt at a Solution I can't seem to understand exactly what I'm supposed to do, what keeps throwing me off is the two different coordinate systems, ij and xy. No matter what combinations of vectors I use, nothing comes...

alright so I can apply conservation of energy by saying: 1/2 Mgh = 1/2*M*v^2 + 1/2 gh * cos(theta) gh = v^2 + gh(cos(theta)) v = sqrt( gh(1-cos(theta)) ) But that would give me an answer of 7.76 m/s^2, which is close, but not right (Webassign gives me an answer of 7.13 m/s^2) *Edit* 1/2 Mgh...

Homework Statement R = 49 m theta = 29° Homework Equations Kinematic equations for Rotational Forces and Circular motion Kinetic Energy = 1/2*I*w^2 The Attempt at a Solution First I can find the moment of Inertia for a rod, which is: 1/12 * MR^2 However I need to use the...

Oh I think I get it so Acc. of A is 3.4883 m/s^2 Acc. of B is 4 m/s^2 F(net) = FA + FB = 24N + 15N = 39N m(net)=mA + mB = 6kg + 4.3kg = 10.3 kg a(net)= F(net) / m(net) = 39 N / 10.3 kg = 3.786 m/s^2 Now I need the force necessary to move A at 3.786 m/s^2...

Alright so to find the net acceleration "a" I set it up like this: ("F" = Net Force, "m" = Net Mass) F=ma 39N = (10.3kg)*a 39N/10.3kg=a a=3.786m/s But this would be the acceleration of the entire system would it not? Doesn't there need to me forces in the opposite direction (-i)...

alright so... acc. of A = 3.4884 m/s^2 acc. of B = 4 m/s^2 but I'm not sure I understand exactly what that means. That the 0.512 m/s^2 difference is somehow related to T? I mean F=ma, but if I were to apply this formula what would be the mass? A? B? the sum of the system?

Homework Statement Here is the problem verbatim (values have been slightly changed, also assume a frictionless environment): "A 4.3 kg block A and 6.0 kg block are connected by a string of negligable mass. Force FA = (15 N) acts on block A; force FB = (24 N) acts on block B. What is...