Question

What is the tension in cord A in this diagram? It would help if you answered in Newtons but not absolutely necessary. I know the answer already (683 N) I would just like to see how to get to the answer

Answer 1

Answer:

683 N

Explanation:

If the block is in equilibrium, then the net force acting on it is zero.

We have therefore to resolve each force along the x- and y- direction, and equate the net force along each direction to zero. Doing so, we obtain the following equations:

$T_B cos 45^{\circ} - T_A sin 60^{\circ} = 0\\T_B sin 45^{\circ} - T_A cos 60^{\circ} - W = 0$

where

$T_A$ is the tension in cord A

$T_B$ is the tension in cord B

W = 250 N is the weight of the block

From the first equation we get:

$T_B = T_A \frac{sin 60^{\circ}}{cos 45^{\circ}}$

And substituting into the second equation, we find the tension in cord A:

$(T_A \frac{sin 60^{\circ}}{cos 45^{\circ}}) sin 45^{\circ} - T_A cos 60^{\circ} - W = 0\\T_A (sin 60^{\circ} tan 45^{\circ} - cos 60^{\circ}) = W\\T_A=\frac{W}{ (sin 60^{\circ} tan 45^{\circ} - cos 60^{\circ})}=683 N$