# B1-2: Tension in a String: Find an Unknown Mass Using Equilibrium of Forces Theory¶

## Apparatus¶

Wooden rod about $$1\text{m}$$ long with eyelets as shown; about $$1.7\text{m}$$ good-quality cord; 2 single pulleys; 2 x $$50\text{g}$$ masses; 5 x $$100\text{g}$$ masses; scale pan; unknown mass; 2 clamps & stands; 2 G-clamps; triple beam balance; metre ruler; 1 sheet graph paper; spirit level.

## Procedure¶

1. Clamp the wooden rod firmly and horizontally, so that there is space for the scale pan and unknown mass to move a large distance vertically without touching any object. Assemble the apparatus as above, placing $$m$$ = $$200\text{g}$$ in the scale pan. Measure and record $$\overline{AB}$$.
2. Move $$m_1$$ up and down, finally placing it in the middle of the range of possible equilibrium positions. Ensure that pulley $$P$$ is directly under the mid-way mark on the rod. Measure and record values of $$m_1$$, $$h$$, and $$d$$.
3. Repeat 2. with $$m_1$$ = $$250\text{g}$$, $$300\text{g}$$, $$350\text{g}$$, $$400\text{g}$$, $$500\text{g}$$, $$600\text{g}$$ each time recording $$m_1$$, $$h$$, and $$d$$. Check that the pulley $$P$$ remains under the mid-way mark on the rod.

## Theory¶

Since point P is in equilibrium (Newton’s 1st law):

$\text{Resultant} \ \tilde{R} = \tilde{T}_1 + \tilde{T}_2 + \tilde{W} = \tilde{0}$

Therefore horizontally:

$\sum{F_x} = 0 \qquad \therefore \ T_2(\cos\theta_2) - T_1(\cos\theta_1) = 0$

However, $$\theta_1 = \theta_2$$ (observation), therefore:

$T_1 = T_2$

And vertically:

$\sum F_y = 0 \qquad \therefore \ T_1(\sin\theta_1) + T_2(\sin\theta_2) - W = 0$

but $$\theta_1 = \theta_2$$ and $$T_1 = T_2$$, so:

$2T_1 (\sin\theta_1) - W = 0 \label{eqn1} \tag{equation 1}$

but $$T_1 = m_1g$$, $$\sin\theta_1 = \frac{h}{d}$$, and $$W = m_2g$$. Hence:

$\frac{h}{d} = \frac{m_2}{2m_1}$

## Analysis¶

1. Plot a graph of $$\frac{h}{d}$$ against $$\frac{1}{m_1}$$, and find the gradient.

2. Use only the gradient and the formula given at the end of the theory to calculate the unknown mass $$m_2$$.

3. Measure the mass of $$m_2$$: on the beam balance, and assuming this is accurate, calculate the % error in the value obtained in (2) above.

1. Use the value of $$\frac{h}{d}$$ when $$m_1 = 400\text{g}$$, to calculate $$\theta_1$$ at this point. Calculate $$W = m_2 g$$.

2. $$m_1$$ is suddenly increased to $$500\text{g}$$...

Assuming that at this moment $$\theta_1 = \theta_2 = \text{the value from (a) above}$$, find the initial upward acceleration of $$m_2$$, as it heads towards a new equilibrium position.

(Hint: find $$T_1$$ and use part of $$\ref{eqn1}$$ above to find the net upward force on $$m_2.)$$