Need help finding tension in string?

Question

A string is attached to three masses 4kg, 1kg and 6kg and the acceleration of gravity is 9.8m/s^2. A pulley has no mass and no friction and the string has no mass. The distance between the 1kg and 4kg is 2m and the string is 24cm long. The tension T1 is between the 4kg and 1kg, and the tension T2 is between 2kg and the pulley, both masses are on the left side and T3 is on the right side with 6kg. The 4kg is below the 1kg on the left side. I need to find the tension between the block with mass of 4kg and 1kg on the left hand side of the pulley and the answer has to be in units of N. Could someone explain how I can answer this.

Answer 1

First, notice that T2 and T3 are the same. Whenever two things are connected directly with a (massless) string over a (massless, frictionless) pulley, the tension is the same everywhere in the string. So that means T2=T3.

Next, just use "F=ma" on each of the masses. And notice that the masses on the left are accelerating up by the same amount that the mass on the right is accelerating down (they have to, since they're tied together). So that means the "a" will be the same amount in all the "F=ma" equations.

Let's use "up" as the positive direction.

For the 6kg mass, use the variables "F6", "m6" and "a6".
The forces on m6 are:
1) weight = (m6)(g) [down]
2) T3 [up]
So:
F6 = (m6)(a6) = T3 − (m6)(g)

For the 4kg mass, the forces are:
1) weight = (m4)(g) [down]
2) T1 [up]
So:
F4 = (m4)(a4) = T1 − (m4)(g)

The 1kg has THREE forces acting on it:
1) weight = (m1)(g) [down]
2) T1 [down]
3) T2 [up]
So:
F1 = (m1)(a1) = T2 − T1 − (m1)(g)

Now we have 3 equations. To help combine them, we note these facts:
1) a1 and a4 are the same ('cause they're connected with a string): a1=a4.
2) a6 is the same as a1 and a4, except it's in the opposite direction: a6=(-a1)=(-a4)
3) T2 and T3 are the same: T2=T3

So, let's rewrite the 3 equations, replacing a4 with a1, a6 with -a1, and T3 with T2:

6kg mass:
(m6)(-a1) = T2 − (m6)(g)

4kg mass:
(m4)(a1) = T1 − (m4)(g)

1kg mass:
(m1)(a1) = T2 − T1 − (m1)(g)

You now have 3 equations in 3 unknowns (a1, T1, T2). From this point you can use algebra to solve for all the unknowns, and in particular T1 which is the quantity you want.

Answer 2

Note there are some discrepancies in your question such as the introduction of the 2kg mass?I will assume that there is a 1 kg and 4 kg mass on one side and 6 kg mass on the other. Note the general procedure I use will work for different system


For each block draw all the forces acting on it. gravity always acts down and Tension always pulls
The 4kg and 1kg accelerate up, the 6kg acc down.

to find the acceleration note that there is 6g pulling on one side and 5g pulling on the other, so the net force on the whole system is
6g-5g = (total mass)a
g = (1+4+6)a, so a = g/11

Now to find tension between 4kg and 1 kg mass
Note there are 2 forces acting on 4kg mass
T up and 4g down, so
applying Fnet = ma to this block alone we get
T- 4g = 4(g/11)
T = 4(1 + 1/11)g = 4(1.0909)g ~ 42.7N

Answer 3

attempt this... import java.util.StringTokenizer; public type TokenizerApp { public static void significant(String[] args) { String fee = "John 23 34 5 6 40 5 3 34"; print(fee); equipment.go out(0); } private static void print(String fee) { StringTokenizer st = new StringTokenizer(fee); String call = st.nextToken(); int entire = 0; at the same time as (st.hasMoreElements()) { entire += Integer.parseInt(st.nextToken()); } equipment.out.format("%s: %dpercentn", call, entire); } }

Answer 4

Spring balance weighting machine.