Gharib

Kevin stands at the edge of a cliff, holding one ball in each hand. He throws one of the balls straight up with speed v, and at the same time he throws the other ball straight down, also with speed v. Ignoring air resistance, which ball hits the ground wi...

A large plate falls to the ground and breaks into three pieces. m1 falls with v1=3m/s, 25 degrees to the left from the vertical. m2 falls with v2=1.79m/s, 45 degrees up from the horizon. And m3=1.3kg falls with v3=3.07m/s vertically down. How would I find...

m=77kg angle=26 degrees Normal force holds half the weight. Positive being up, Y direction: (1/2)w-(F1)cos(26)-(F2)cos(26) =(1/2)(77kg)(9.8m/s^2)-2Tcos(26) [since T1=T2 => T=377.3N/2cos(26) T=210N

m=77kg angle=26 degrees Normal force holds half the weight. Positive being up, Y direction: (1/2)w-(F1)cos(26)-(F2)cos(26) =(1/2)(77kg)(9.8m/s^2)-2Fcos(26) [since F1=F2 => F=377.3N/2cos(26) F=210N

M=0.393kg m=0.181kg Draw the FBD, you will see that T1+T2-W(of M)=0 Eq 1 T2-W(of m)=0 Eq 2 (T2 is what we need) now, Eq1-Eq2 => T1+Wm-WM=0 T1= -g(m-M) T1= x Newtons now, plug in T1 in Eq 1, T2=WM-T1 *Reminder W=mass*9.8 Ans: T2=1.77N

Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method]. In that case, Posit...

Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method]. In that case, Posit...

Tilt your axis such that, Y is pointing to the opposite of Weight of M. And X is pointing towards the right. (like usual). [The axis shown in the diagram requires you to find Fnet, which I don't know how to find; thus use this method]. In that case, Posit...

If you draw the FBD for both object separately. Then, Positive being up and right, T=m1a Eq 1 T-m2g=-m2a Eq 2 T= -m2a+m2g Eq 3 Eq 3 = Eq 1 m1a= -m2a+m2g a=m2g/(m1+m2) since, a=x then, T=m1x

You have to make two FBD; 1st for the force being applied to the leg, and 2nd for the mass hanging from cable. Positive being up and right. Y direction for FBD #2: T= mg Eq 1 x direction for FBD #1: -F+2Tcos(60)=0 Eq 2 T= F/2cos(60) T=x From Eq 1, m=x/g t...

Is there any angle with the horizon? if I am not mistaken, then T= ma

@risha, equate (1/2)mv^2=8.5F m and v are known, solve for F, therefore F=604N

Thanks Bobpurley, I didn't know it was that simple.