Simple to some but !!!

[Gareth56]

I've hit a brick wall in trying to see how in a text book they went from:-

 

T1xsin37deg + (1.33T1)(sin53deg) -100N = 0

 

straight to

 

T1 = 60.1N

 

 

 

Any help on the intervening steps would be appriciated.

 

Thanks

[timo]

You rearrange the equation for T1 (i.e. in the form T1=....), then type the right-hand side into the calculator of your choice. Note that XY + XZ = X(Y+Z), in case your problem is rearranging.

[Gareth56]

Thanks for the pointer but sadly what's throwing me is the (1.33T1)sin53 term I just don't know what to do with that.

 

The closest I've got is T1 = 51.7N

[Fuzzwood]

Simple: 1.33xsin53xT1, and sin53deg is ofc just a simple number which you can work out first.

Edited July 22, 2009 by Fuzzwood

[the tree]

Here it is step by step, whereabouts are you struggling?[math]\begin{array}{ccc}

T_1 \sin{37} + 1.33 T_1 \sin{53} - 100 N & = & 0 \\

T_1 \sin{37} + 1.33 T_1 \sin{53} & = & 100 N \\

T_1 ( \sin{37} + 1.33 \sin{53} ) & = & 100 N \\

T_1 & = & (\frac{100}{ \sin{37} + 1.33 \sin{53} }) N \\

T_1 & \approx & 61 N \\

\end{array}[/math]

 

a simple number which you can work out first.

Evaluating floating points anywhere before the last stage means that you risk accumulating unnecessary rounding errors - avoid this.

[Gareth56]

I thought that when you got (1.33T1)sin53 which I presume is the same as (1.33 x T1)0.8, then everything inside the bracket had to be multiplied by everything outside the bracket.

[Fuzzwood]

I thought that when you got (1.33T1)sin53 which I presume is the same as (1.33 x T1)0.8, then everything inside the bracket had to be multiplied by everything outside the bracket.

 

The ( ) are not needed in that part, as you simply are multiplicating a multiplication of 2 other terms. You will need to do that however, when it would have been (1.33 + T1)0.8

[the tree]

I thought that when you got (1.33T1)sin53 which I presume is the same as (1.33 x T1)0.8, then everything inside the bracket had to be multiplied by everything outside the bracket.

Yeah, but multiplication (as far as numbers are concerned) is associative, meaning that [math](a\times b)\times c[/math] and [math]a \times ( b \times c ) [/math] and [math] a \times b \times c[/math] all amount to the same thing.