A stake is being pulled out of the ground by means of two ropes as shown, Knowing that the tension in one rope is Problem 1.- A stake is being pulled out of the ground by means of two ropes as shown, Knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N

Question-Answer › Category: Engineering Mechanics › A stake is being pulled out of the ground by means of two ropes as shown, Knowing that the tension in one rope is Problem 1.- A stake is being pulled out of the ground by means of two ropes as shown, Knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N

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GME asked 11 months ago

Problem 1.- A stake is being pulled out of the ground by means of two ropes as shown, Knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N

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Mazurek Gravity answered 11 months ago

Given data :

Force by rope 1 , = 120 N at angle of 25° from vertical axis in anticlockwise direction
Force by rope 2, = P Newton at angle of $\alpha$ ° from vertical axis in clockwise direction
Resultant force , R = 160 N in verticle upward direction

Now, as shown in figure we can get two component of F_1 and  , in vertical and horizontal direction.
So, from figure we say that,
Resultant force in horizontal direction, $F_H= Psin\alpha-120sin25^o$
But it is given that, resultant force must be in vertical dorection only. So, FH = 0
So, $Psin\alpha-120sin25^o= 0$
So,   $Psin\alpha=120sin25^o$
So,   $Psin\alpha=50.71419141\ N$ ________(1)
Same way, Resultant force in vertical direction, $F_V= Pcos\alpha+120cos25^o$
We have given resultant force in vertical direction i.e. 160 N so, $F_V= 160 \ N$
So, $Pcos\alpha+120cos25^o= 160$
So, $Pcos\alpha= 160-120cos25^o= 51.24306556\ N$ _______(2)
Now taking square of equation (1) and equation (2) and then adding them ,
$(Psin\alpha)^2+(Pcos\alpha)^2= (50.71419141)^2+(51.24306556)^2$
So, $P^2[(sin\alpha)^2+(cos\alpha)^2]= 5197.780978$
Now, $(sin\alpha)^2+(cos\alpha)^2 = 1$ ,
So,
So, $P= 72.09563772 \ N\ \approx \ 72.1\ N$
Now, from equation (1),
$Psin\alpha=50.71419141\ N$
So,   $72.09563772\times sin\alpha=50.71419141$
So, $sin\alpha=0.703429403$
So, $\alpha=sin^{-1}(0.703429403)= 44.70279717^o$
So, Force by rope 2 is 72.1 N and angle $\alpha$ is equal to 44.7° .