# Frictionless pulley problem

1. Oct 27, 2004

### SnowOwl18

I'm trying to help a friend with this problem and we are both very stuck. I wanted to ask for help here for him.

-------A 1.0 kg tin of anti-oxidants on a frictionless inclined surface is
connected to a 2.0 kg tin of corned beef. The pulley is massless and frictionless. An
upward force F r of 6.0 N acts on the corned beef tin, which has a downward acceleration
of 5.5 m/s2. (a) What is the tension in the connecting cord? (b) What is the angle b ?--------

Unfortunately, I can't post the picture of the problem...but basically there is one tin (anti-oxidants) on an incline and attached to it is a cord that goes through a pulley...and on the other end of the cord, hanging straight down is the other tin (corned beef). The angle, b, is the angle of the incline, i think. I found an equation that might work for finding the tension. T= [(m1m2(1+sintheta))/ m1 + m2]g . But the problem is we don't know the angle or how to find it. If anyone could please help us out, we'd greatly appreciate it.

2. Oct 28, 2004

### maverick280857

Firstly, your answer for T does not include the upward force Fr acting on the hanging mass. Hence it is incomplete.

In my solution below, $$m_{1}$$ denotes the mass on the incline and $$m_{2}$$ denotes the hanging mass

The equations of motion are:

$$T - m_{1}g\sin\alpha = m_{1}a$$ (for the mass on the incline)
$$m_{2}g - T - F_{r}= m_{2}a$$ (for the hanging mass)

Adding these two equations, you get

$$m_{2}g - m_{1}g\sin\alpha - F_{r}= (m_{1}+m_{2})a$$

You can see for yourself that both the masses are constrained to move with the same (magnitudinally) acceleration for the rope to remain taut. The acceleration is given to you so you can go ahead and solve for the angle alpha from this equation. From the second equation, you can solve for T.

Hope that helps.

Cheers
Vivek
(I hope I've understood your question correctly)

3. Oct 28, 2004

### SnowOwl18

Thanks so much! )