physics machine problems?

Question

uhhhh what? i'm have no clue how to find these equations...can someone show me step by step so that i will know how to do them later...thanks in advance!!

1. Calculate the acceleration and the tension in the rope. For your variables use "m1", "m2", "theta1", "theta2" and "g". Assume all surfaces are frictionless, and the pulley and rope are massless and frictionless.

Diagram: http://www.webassign.net/userimages/48539?db=v4net

2. Find the acceleration for each of the following machines (use the same notation conventions as the previous problems; i.e. m1, m2, theta1, theta2, g, etc.).

Diagram 1: http://www.webassign.net/userimages/48540?db=v4net

Diagram 2: http://www.webassign.net/userimages/48541?db=v4net

Diagram 3: http://www.webassign.net/userimages/48542?db=v4net

Answer 1

1. The way to approach these is to choose a direction of motion and set that as the positive. Typically to the right is positive.
Next, use F=m*a and calculate the net force.

For m2:
m2*g*sin(th2)-T=m2*a
where T is the tension in the rope.

Since the rope connects the two blocks and there is no friction in the system, then a is equal for both.

for m1:
T-m1*g*sin(th1)=m1*a

now you have two equations and two unknowns, a and T:
T-m1*g*sin(th1)=m1*a
m2*g*sin(th2)-T=m2*a
do some algebra

T=m1*(a+g*sin(th1))
T=m2*(g*sin(th2)-a)

subtract
m2*(g*sin(th2)-a)=
m1*(a+g*sin(th1))

isolate a
a=g*(m2*sin(th2)-
m1*sin(th1))/(m1+m2)

once you solve for a, you can then solve for T

for this question, plug the a into the equations for T and do some more algebra.

2) More of the same. I will give you equations for diagram 2:

T1-m3*g=m3*a
T2-m1*g*sin(th1)-T1=m1*a
T3+m2*sin(th2)-T2=m2*a
m4*g-T3=m4*a

there is only one a, and three tensions. You have 4 equations, so you just do a bunch of algebra to find a and all of the tensions.


j

Answer 2

Power = paintings / time paintings = drive x distance The enter energy is already given to you, that's the 18 hp motor. 18 hp changed to kilowatts is thirteen.forty two kilowatts.So the enter energy may be thirteen,420 watts. a) paintings = drive x distance mg = 3000 kg * nine.eight m/s^two = 29,four hundred Newtons of drive knowledge power = KE = paintings output = mgh = mass * gravity * peak paintings = 29,400N * eight.0m = 235,two hundred N*m = 235,two hundred Joules of labor output b) energy enter = thirteen.forty two kw energy output = paintings / time energy output = 235,two hundred J / 20 seconds = eleven,760 watts of energy output So the 18 hp engine (thirteen,420 watts) offers output energy of eleven,760 watts. So the potency is the output energy divided via enter instances one hundred: c) eleven,760 watt / thirteen,420 watt = zero.88 * one hundred = 88 % effective for the hoist and engine