Show that the force applied by each rope

[Gughanath]

please help!

i don't know how to solve da problem please help:

Two ropes are used to carry a load of weight 400N. The ropes are 40 degrees to the horizontal. Show that the force applied by each rope is to the container is abou 300N.

 

[mrjeffy321]

in order to hold up an object weighing 400 Newtons, you must apply a force of 400 Newtons upward to counter act the force of weight.

if you have two ropes, at 40 degree angles, holding it up, then you know that the combined upward force those ropes are exerting on the object must be 400 Newtons.

also, if you kow the "resultant" force in each rope, the force along the hypotenuse of the imaginary triangle, and the angle of each, then you can calculate what the ropes' upward forces are,
Fy = F * sin(angle), where Fy is the upward force, F is the total resultant force.
so it is 300 * sin(40) = 193 Newtons upward, and there are 2 ropes exerting the same force, so the total upward force is 386 Newtons. this isn't quite the 400 Newtons that we were looking for, but is close enough I think sonsidering they only used 1 sig. fig in the question, then 1 sig fig in the answer would be 400 Newtons.

 

[PBRMEASAP]

i don't know how to solve da problem please help:

Two ropes are used to carry a load of weight 400N. The ropes are 40 degrees to the horizontal. Show that the force applied by each rope is to the container is abou 300N.

By symmetry of the problem (I'm assuming it is because you didn't say it wasn't), the tension in each rope provides 200N of force in the upward direction. This upward force can also be written in terms of the tension T in the rope:

$$\mbox{upward force} = T \sin 40^{\circ} = 200N$$

Now you can solve for T, which is what you are looking for.

 

[vitaly]

Here's my guess:

$$-T_1\cos(40)+T_2\cos(40)=0<br /> <br /> T_1\sin(40)+T_2\sin(40)-400N=0$$

Those two are equations above are for your x- and y-coordinates. The force must equal zero.

The first equation simplifies into:

$$T_1=T_2\cos(40)/\cos(40)$$

*T just stands for tension of cables 1 and 2

Now plug in $$T_1$$ into the second equation.

$$T_2\cos(40)\sin(40)/cos(40) + T_2\sin(40)=400 N<br /> <br /> = T_2\left(\cos(40)sin(40)/cos(40)\right)+sin(40)=400 N<br /> <br /> = T_2\left(1.28\right)=400 N<br /> <br /> = T_2=312.5 N$$

Now you know tension of one cable, then just plug it in your original equation:

$$T_1=T_2\cos(40)/cos(40)<br /> <br /> = T_1=312.5 N$$

They equal about 300 N, like the problem says.

*I'm new with the latex equation stuff, so I'm sorry if these equations have mistakes.