The system in the figure below is in equilibrium, with the string in the center exactly horizontal. Block A weighs 37.0 N, block B weighs 47.0 N, and angle φ is 37.0˚. What are (a) tension T1, (b) tension T2, (c) tension T3, and (d) angle θ. ?

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The system in the figure below is in equilibrium, with the string in the center exactly horizontal. Block A weighs 37.0 N, block B weighs 47.0 N, and angle φ is 37.0˚. What are (a) tension T1, (b) tension T2, (c) tension T3, and (d) angle θ. ?

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Answer 1 · 2015-09-25T19:09:47

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$T 1 = 46.3$ $T1 = 46.3$ $N$ $N$
$T 2 = 27.9$ $T2 = 27.9$ $N$ $N$
$T 3 = 54.6$ $T3 = 54.6$ $N$ $N$
$θ = {30.7}^{o}$ $theta = 30.7^o$

Explanation:

The fact that the string in the center of the figure is horizontal means it is providing no vertical force. All the tension that holds up each block comes from the angled strings at the left and right.

The string at the left, which we can call 'string 1' and has tension $T 1$ $T1$ , is holding up all of the weight of block A, and the string to the right, 'string 3' with tension $T 3$ $T3$ is holding up block B.

For the blocks to be stationary, all the forces on them must sum to 0, so the upward force supplied by the string is just the same as the weight force of the block, which we are given. We need to find what the total tension in string 1 needs to be in order to have a vertical component of $37$ $37$ $N$ $N$ . $T 1$ $T1$ is the hypotenuse of a right-angled triangle, and the opposite side to the angle $ϕ$ $phi$ ( $37^{o}$ $37^o$ )is the horizontal force, the adjacent side is the vertical force. We know from trig that:

$cos θ = \frac{adjacent}{hypotenuse}$ $costheta=("adjacent")/("hypotenuse")$

Rearranging and substituting in:

$T 1 = \frac{vertical force}{cos ϕ} = \frac{37}{cos 37} = 46.3$ $T1 = ("vertical force")/(cos phi) = 37/(cos 37) = 46.3$ $N$ $N$

Using the same kind of reasoning, the horizontal force on this string, which is equal to $T 2$ $T2$ will be:

$T 2 = T 1 sin ϕ = 46.3 \cdot sin 37 = 27.9$ $T2=T1sinphi = 46.3*sin37=27.9$ $N$ $N$

This will also be the horizontal component of the force provided by string 3 on the second mass:

$sin θ = \frac{T 2}{T 3}$ $sintheta = (T2)/(T3)$ : call this Equation 1

We know $T 2$ $T2$ but don't yet know either $θ$ $theta$ or $T 3$ $T3$ . We can use the vertical force on block B to find both. We do know that the weight of block B is $47$ $47$ $N$ $N$ .

$T 3 = \frac{vertical force}{cos θ} \to T 3 = \frac{47}{cos θ}$ $T3 = "vertical force"/(cos theta) to T3 = 47/cos theta$

Substitute this value for T3 into Equation 1:

T2/( $\frac{47}{cos θ}$ $47/cos theta$ ) = $sin θ$ $sin theta$

Invert and multiply:

T2 x $\frac{cos θ}{47}$ $(cos theta)/47$ = $sin θ$ $sin theta$

Substitute in the known value of T2 and rearrange:

$\frac{sin θ}{cos θ}$ $(sin theta)/(cos theta)$ = $\frac{27.9}{47}$ $27.9/47$

$tan θ$ $tan theta$ = $\frac{27.9}{47}$ $27.9/47$

$θ = {tan}^{- 1} (\frac{27.9}{47}) = {30.7}^{o}$ $theta = tan^-1(27.9/47) = 30.7^o$

We can then use this in Equation 1 to find the value of T3:

$sin θ = \frac{T 2}{T 3}$ $sintheta = (T2)/(T3)$

Rearranging and substituting in the value of T2:

T3 = $\frac{27.9}{{sin 30.7}^{o}}$ $(27.9)/(sin 30.7^o)$ = 54.6 N

Summing up, then:

$T 1 = 46.3$ $T1 = 46.3$ $N$ $N$
$T 2 = 27.9$ $T2 = 27.9$ $N$ $N$
$T 3 = 54.6$ $T3 = 54.6$ $N$ $N$
$θ = {30.7}^{o}$ $theta = 30.7^o$

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The system in the figure below is in equilibrium, with the string in the center exactly horizontal. Block A weighs 37.0 N, block B weighs 47.0 N, and angle φ is 37.0˚. What are (a) tension T1, (b) tension T2, (c) tension T3, and (d) angle θ. ?

1 Answer

Explanation:

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