Force to tension strings

Question

The graph shows a block of mass mkg that is at rest. Determine the strength modulus with which the rope 3 must be pulled so that the other two ropes are tight. (Consider $θ = 30º; g = 10m / s²$)

answer: 40N

The part F = 80N I found, but why F = 40N?

Answer 1

You need to write the sum of the forces along the two axes. Say tension in rope 1 is $T_1$ and in rope 2 is $T_2$. Then, in the vertical direction: $$F\sin\theta+T_1\sin\theta=mg$$ and in the horizontal direction $$F\cos\theta=T_1\cos\theta+T_2$$ You have two conditions for the ropes to be tight: $T_1\ge0$ and $T_2\ge0$. Looking at the first equation, the maximum force occurs when $T_1=0$, so $$F_{max}=\frac{mg}{\sin\theta}=\frac{4\cdot 10}{1/2}=80N$$ We now rewrite the first equation as $$F=\frac{mg}{\sin\theta}-T_1$$This means that as $T_1$ increases, $F$ will decrease. But how far can we go? We rewrite equation 2 as $$F=T_1+\frac{T_2}{\cos\theta}$$ For a fixed $F$, you can increase $T_1$ and decrease $T_2$. at the maximum $T_1$ you have $T_2=0$. So in that case $F=T_1$. Plugging back into the first equation $$2F\sin\theta=mg$$which for the given $theta$ means $$F_{min}=mg=40N$$

Answer 2

Here is a shortcut using the fact that the block is in equilibrium and that the forces $F_1$ and $F_3$ along rope 1 and 3 are symmetric wrt. the vertical axis through the center of the block.

Let $v_1$ and $v_3$ be the magnitude of the vertical components of $F_1$ and $F_3$. Then you have

• $v_1 = v_3$ because of above mentioned symmetry.
• $v_1 + v_3 = 2v_3 = 40N \Rightarrow v_3 = 20N$.
• $\frac{v_3}{F_3}=\sin 30^° =\frac 12$.

Hence,

$$F_3 = 2v_3 = 40N$$