# What is the minimum tension (in Newtons) in the angled rope?

## Homework Statement

A hanging 2.00-kg (19.6N) block begins to accelerate a 4.28-kg (41.94N) block on a horizontal surface. But a rope connected to the wall at 26.9 degrees brings the block to an equilibrium position. The coefficient of friction between the surface and the 4.28-kg (41.94N) block is 0.162.

What is the minimum tension (in Newtons) in the angled rope that maintains this equilibrium condition?

The answer is 15.6 Newtons but I cannot figure out how to get there.
Here's a picture: http://gbhsweb.glenbrook225.org/gbs/science/phys/chemphys/audhelp/u8setf/q14.gif

## Homework Equations

T-mgsinθ=ma so T=ma+mgsinθ
F=ma

## The Attempt at a Solution

I've tried many different ways. I've tried to find both the horizontal and vertical forces of the rope and picturing it as a triangle. I may just be adding the forces wrong or using the wrong equations. I found that the frictional force of the 41.94N block is 6.79N. I thought that the tension of the rope between the two blocks is 19.6N (equal to the weight of the hanging one). I thought I had found the answer when I subtracted 6.79+19.6 from 41.94 (equals 15.55). But I know that isn't the right way to do it. Help please.

Simon Bridge
Homework Helper
Show us your working one step at a time ... do not substitute any numbers until all the algebra is finished.

Most likely mistakes include:
getting the trig wrong for the rope to the wall.
forgetting the upwards pull of that rope when calculating friction.

arildno