# Find force of friction on an object under influence of tension by another object

I conducted an experiment using video analyses software, and found the accleration of a box on a table being pulled by a string attached to a pulley on the edge of the table attached to the weight (bag of sand). (the meter stick is used for the video analysis reference)

•The mass of the sand is 9.883Kg, and the mass of the box is .3636Kg

•I am using the value 9.8m/s^2 for the acceleration of gravity.

•The acceleration of the box was calculated to be 6.034m/s^2 using the software.

•I am trying to find the force of friction.

So I tried a few things, but I don't know if they are correct, and would appreciate any help with the problem.

I first calculated the force Gravity is pulling on the bag of sand. 9.8m/s^2 * 9.883 Kg = 96.8534 Newtons. I then calculated the Effective force on the sand, 6.034 m/s^2 * 9.883 Kg = 59.594 Newtons. That would mean 96.8N-59.495N = 39.259 Newtons of friction force on the sand.

But the box's mass has something to do with it too, correct? I did the same calculation for the wooden box, and got 3.563N from gravity, 2.193N Effective, and 1.37N of friction.

I am confused here, do I add the friction forces together? It would seem like there should only be one friction force from the box. Everything is held together by the string, so I would think it would have the same value throughout.

Thanks for any help, I really wish to have a better understanding of newtons laws and motion.

The tension pulls the box to the right, and friction to the left. The net result is horizontal acceleration

$$T - \mu \,m_{box} g = m_{box} a_{box}$$

The tension is found by the acceleration of the bag of sand

$$m_{bag} g - T = m_{bag} a_{bag}$$

Since they are connected with a rope $a_{bag} = a_{box}=6.034$. Now you have a system of equations

\begin{aligned} T - \mu\,(0.3636*9.8) & = (0.3636*6.034) \\ (9.833*9.8)-T &= (9.833*6.034) \end{aligned}

which you need to solve for $T$ and $\mu$.

I do apologize for being a little to artistic: Force of Sandbag $F_s$, Normal Force $F_n$ (equal to Gravitational Force $F_g$), and Kinetic Friction $F_k$: $$F_s = 9.883 \cdot 9.8 \\ F_s = 96.8534\space N$$ We also know the total accelaration which means we know the Net Force: $$\Sigma F = ma \\ \Sigma F = (0.3636 + 9.883) \cdot 6.034\\ \Sigma F = 61.8279844\space N\\$$ Since we know the $F_s$ and the Net Force, we can easily calculate the kinetic force: $$F_k = F_s - \Sigma F \\ F_k = 96.8534-61.8279844 \\ F_k = 35.0254156\space N$$ From here we can find the coefficient of kinetic friction: $$F_k = \mu F_n \\ F_n = \frac{F_k}{\mu} \\ 3.56328 = \frac{35.0254156}{\mu} \\ \mu = \frac{35.0254156}{3.56328} \\ \mu \approx 9.83\\$$ I included the mass of the sandbag in the Net Force. Honestly, I'm not positive I should have. However, I only calculated the Kinetic Frictional Force of the box.

• 9.8 seems pretty high for coefficient, and 35 newtons seems like a lot of force for a small box Oct 17, 2014 at 16:12
• Oh, dumb me, the mass was 1 kg, not 10 Oct 17, 2014 at 16:17