Recent content by ndwiseguy

perfect thank you very much

Fpcos(x)-(0.217*(Fpsin(x)-m1g))-(m2g+m2a)=m1a Look right?? The final answer i got was 1.82 m/s^2

The only other force I can see acting normal to the friction surface is the y-component of the pulling force. So would I use both that and mg??

so my two equations are: Fpcos(x)-Ff-T=m1a T-m2g=m2a Now i solved eqn two for T and got T=m2a+m2g which i then inserted into the first question. Now, for the force of friction, would i use 0.217*9.8*33.1 because the normal force would be equal to the force of gravity? If so, my answer is...

A 33.1- kg block (m1) is on a horizontal surface, connected to a 5.7- kg block (m2) by a massless string. The pulley is massless and frictionless. A force of 236.1 N acts on m1 at an angle of 30.9 °. The coefficient of kinetic friction between m1 and the surface is 0.217. Determine the upward...

thank you very much

Ok now, i know it says that i should show the work I've done on the question. Problem is, I have no idea where to start. So here is the question: Assume a cyclist of weight w can exert a force on the pedals equal to 0.90w on the average. If the pedals rotate in a circle of radius 18cm, the...