Right, and you know that mu_s and mu_k are related by the equation F(friction)=mu*(F(normal))
and because this is a frictionless, massless pulley we are able to relate the tensions of the two boxes.
The sum is the addition of the i, j, and k components of the vector. To find the magnitude of the sum you first need to add the components then find the magnitude of the resultant vector. i.e. take the square root of the sum of the squares.
Ok Ill concede. But I was just trying to avoid directional confusion. Since the problem starts with an - direction for acceleration, and to simplify it I did the problem assuming a positive acceleration.
Edit* there it should be fixed.
What do you mean there is no deceleration? If the force applied produces a larger acceleration than is present. Then that means in effect the force of friction must be slowing the object down or decelerating it.
Is there a better way you could put it? Sometimes I'm a little confusing.
Well, not quite, the force applied is not the same as the force of friction because there is an acceleration. The only time the force of friction is the same as the force applied is when acceleration is zero.
The reason they give you an initial force is so you can find the -acceleration due...
Not quite,your equation says deltax=.5(v+vo)t
so delta x is a change in x or xfinal - xinitial so runnerA:x-3mi=.5(8mi/h)t
runner B:x-1mi=.5(5mi/h)t
so you have two equations with two unknowns solve for t you should get the same value of t for each runner.
Once you find t you can find the...
You should not have two values for time here. This isnt possible, how can 2 runners meet at different times if they are running towards each other. They should only meet once so you should have one value for t.
Lets break this down. The problem gives you a distance east of the flagpole for...
Well here is a general form of the equation.
M_{1}V_{1x}+M_{2}V_{2x}=M_{1}V'_{1x}+M_{2}V'_{2x}
and another equation for the y components
M_{1}V_{1y}+M_{2}V_{2y}=M_{1}V'_{1y}+M_{2}V'_{2y}
but in the problem they give you an angle that the first ball deflects at being 29.7 degrees below the x axis...
The equation you are using is right but you will end up with two equations.
The mass is the same no matter which direction so all you are concerned with is the velocity which moves with different x and y components depending on the angle. So just use sin's and cos's to find these components...
Well I am going to assume two things so tell me if I am wrong. The mass of both balls are the same and that you know the equation for momentum.
The way you start this problem and all problems like it is to break this up into components, find the sum of the momentums in the x direction and...
im confused about the last part of your explanation when you multiplied the 300 by (4.5-x) why by do you take half the length of the beam and subtract that from x...sorry i cant explain why i dont understand this in more detail.
1) draw a free body diagram of the problem
2)find the force of gravity on the cart,
3)and add this to other forces in that direction, you know the true acceleration therefore use newtons second law to find force,
hope this helps :)
quick question,
A uniform, aluminum beam 9.00 m long, weighting 300 N, rests symmetrically on two supports 5.00 m apart. A boy weighing 600 N starts at point A and walks toward the right.
How far beyond point B can the boy walk before the beam tips?
ans=1.25 m
How far from the right end of the...
oh i get what your saying, sorry thats what it was meant to signify. one last question it gives me the weight of the door in Newtons however i dont think that this is needed so do i just solve for the mass.
i guess what i am trying to say is the weight is unnecessary for the problem right...
the ball bounced and is now travelling in the opposite direction therefore the angular momentum would be negative
L_ball(beforecollision) + 0 = -L_ball(aftercollision)+L(of bar)
oh right the units agree and would be
L(before collision)=m(kg)*10(m/s)*1.5(m)
L(after collision)=m(kg)*6(m/s)*1.5(m)
so then how can i relate this to the angular momentum of the bar
would i add these together because they act in the same direction and then set them equal to the angular...
hmm, im not quite sure i understand exactly what you are saying. because before the collision the angular momentum of both the bar and ball would be 0 since it does not state that the ball is spinning. sorry im a little confused on what you mean
Hi i have a question with this problem
A thin, uniform, metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball...
Calculate the net torque about point O for the two forces applied as in the figure View Figure . The rod and both forces are in the plane of the page. Take positive torques to be counterclockwise.
A image is here to show the problem
What i did
12Nsin(30)*(2m)-(8N*3m)
or if you like this...
Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge.
I know that I=n*m*r^2
where n is the inertial constant
but i think my main problem with this is where the axis of rotation is, im thinking...