Recent content by aliciaw0

do u mean [5,0] and [0,16] because i used vectors and i got the angle of 72.6 deg. and that was wrong

oh ok.. so i redid it and i got m2 y: 2T-m_2*g=m_2*a_2 since T= m_1*a_1 i subsituted that it and solved for a_1 and got a_1= (2*m_2*g)/(4m_1+m_2) is this right?

The paper delivery boy tries to throw the paper into your narrow driveway without slowing down. His pickup truck travels at 5.00 mph, and he throws the paper at 16.0 mph just as the truck passes the driveway. in what direction should he throw the paper in order for it to land in the driveway...

x m_1: T=m_1*a_1 y m_2: T-m_2*g=m_2*a_2 2(m_1*a_1)*g=m_2*a_2 a_2= a_1/2 2 (m_1*a_1)*g=m_2*(a_1/2)

oh that's right, it never did say anything about it being frictionless, maybe that's what i did wrong. and there was no other information about the masses standing still

In the figure, find an expression for the acceleration of m_1 i don't understand why (-m_2*g)/(((-m_2)/2)-m_1) isn't right because the a of mass#2 is (-a_1)/2 right?

yepp yepp i had to use sin to get it right so my final equation did turn out to be 2Tsin(8)=F+Fg and that worked. Thanks! =]

is this right? x: Tsin(theta)-Tsin(theta) y: 2Tcos(8)=F+Fg

A 68.0 Kg tightrope walker stands at the center of a rope. The rope supports are 10 m apart and the rope sags 8.00 degrees at each end. The tightrope walker crouches down, then leaps straight up with an acceleration of 8.10 m/s^2 to catch a passing trapeze. What is the tension in the rope as...

yay i got it right thankkk you Y: Fn= (m1*g)/cos(theta) X: (m1*g*tan(theta))/m1 = a system: F= (m1+m2) ((m1*g*tan(theta))/m1 ) =]

thanks i tried what you told me too and i think i still might have done it wrong? f(block1)y: Fn-Fgcos(theta)=0 F(block1)x: Fgsin(theta)=m1a Fgsin(theta)/m1= a system x: F= (m1+m2) ((m1g*sin(theta)/m1) and they said my answer was wrong because it doesn't depend on theta...

Find an expression for the magnitude of the horizontal force F in the figure for which m1 does not slip either up or down along the wedge. All surfaces are frictionless. since it has nothing to do with the friction i am thinking that m1 and the force have to have the same acceleration in the...

i don't know that's exactly what i did and handed to my prof and my answers were right =] haha so i don't know at least its right? =] thanks though

o man lol i made it SO much harder then it really was. lol well for anyone else struggling EVER with this problem like EVERYONE else said there are two components to the acceleration in this problem. a= sqrt( a(radial) + a(tangetial) ) the tangential is given by the foward force. Which is...

can someone let me know if I am getting close to the right answer? so far i have forces in the y= Fn-Fg=ma=0 Fn=Fg forces in the x= F(foward)-F(friction)=ma=mv^2/r 1000N-(u(static)*F(normal))=1900v^2/20 to finish the problem i solve for V and then divide...