Question

A 1200 kg steel beam is supported by two ropes. Each rope has a maximum sustained tension of 6400 N.. . Does either rope break? If so, which one(s)?. Rope 1, rope 2, or none?. . What is the tension in each rope? . rope 1 (in N). . rope 2 (in N).

Answer 1

The weight of the object is
$W=mg=(1200 kg)(9.81 m/s^2)=11172 N$

Now we have to write the equations of equilibrium on both axis, horizontal and vertical one. Calling T1 the tension of the rope on the left, T2 the tension of the rope on the right, the two equations are:
x-axis: $-T_1 \sin 20^{\circ}+T_2 \sin 30^{\circ}=0$
y-axis: $T_1 \cos 20^{\circ}+T_2 \cos 30^{\circ} - W=0$

Let's solve the system. From the first equation we find:
$T_1 = \frac{\sin 30^{\circ}}{\sin 20^{\circ}}T_2$
and if we substitute into the second equation, we get:
$\frac{\sin 30^{\circ}}{\sin 20^{\circ}} \cos 20^{\circ} T_2 + \cos 30^{\circ}T_2 =W$
$T_2= \frac{W}{ \frac{\sin 30^{\circ}}{\sin 20^{\circ}} \cos 20^{\circ} +\cos 30^{\circ}} = \frac{11172 N}{2.24}=4988 N$ --> this is the tension in the second rope

And by substituting into the first equation, we find:
$T_1 =\frac{\sin 30^{\circ}}{\sin 20^{\circ}}T_2=\frac{\sin 30^{\circ}}{\sin 20^{\circ}}(4988 N)=7292 N$ --> this is the tension in the first rope

The maximum tension the two ropes can sustain is 6400 N: therefore, from the value of $T_1$ and $T_2$, we see that the first rope won't break, while the second rope will break.

Answer 2

The tension in first rope is $\boxed{7292\text{ N}}$ while tension in second rope is $\boxed{4988\text{ N}}$. The first rope will break while second rope won’t break as the tension in first rope exceeds the maximum permissible value.

Further explanation:

The maximum permissible tension for both the ropes is 1600 N. so if the tension in either rope exceeds the maximum permissible limit, it will break. The tension in either rope can be calculated by using force equilibrium.

Given:

The mass of the steel beam is $1200\text{ kg}$.

Concept:

Let the tension in the first rope is ${T_1}$ and the tension in the second rope is ${T_2}$.

The weight of the steel beam is

$\begin{aligned}W&=mg\\&=({1200\,{\text{kg}}})({9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}})\\&=11772\text{ N}\end{aligned}$

Draw free body diagram for steel beam as attached below.

Resolve component of ${T_1}$ and ${T_2}$ in horizontal direction and use equilibrium condition.

$\begin{aligned}\sum\limits_{}^{} {{F_x}&=0}\\- {T_1}\sin 20 + {T_2}\sin 30&=0\\{T_1}&=\frac{{\sin 30}}{{\sin 20}}{T_2}\\\end{aligned}$

Simplify the above expression we get

${T_1} = 1.46 \,{T_2}$                              …… (1)

Resolve component of ${T_1}$ and ${T_2}$ in vertical direction and use equilibrium condition.

$\sum\limits_{}^{} {{F_y}&=0}\\{T_1}\cos 20 + {T_2}\cos 30&=W\\0.939{T_1} + 0.866{T_2}&=11772$

Substitute value of ${T_1}$ in the above expression.

$0.939\left( {1.46{T_2}} \right){\text{ }} + {\text{ }}0.866{T_2}{\text{ }}&={\text{ }}11772\\2.236{T_2}&=11772\\{T_2}&=5264.7\,{\text{N}}$

Substitute $5264.7\text{ N}$ for ${T_2}$ in equation (1).

$\begin{aligned}{T_1}&=1.46\left( {5264.7\,{\text{N}}}\right)\\&=7686.46\,{\text{N}}\\\end{aligned}$

From the above calculations it can be observed that tension in $rope\;1$ is more than the maximum permissible limit but tension in $rope\;2$ is less than the maximum permissible limit, therefore $rope\;1$ will break while $rope\;2$ won’t break as the tension in $rope\;1$ exceeds the maximum permissible value.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Laws of motion

Keywords:

1200 kg. Steel beam, supported, two ropes, maximum, sustained, tension, 6400 N, either, break, rope1, rope2, none, 5264.7 N, 7686.46 N.