The tension in the string is-600 N80 N1200 N800 N

Question

Toppr Admin · Accepted Answer

For equilibrium-xA0-about P-xA0-x2211-x3C4-p-0or-500-1-80-0-5-x2212-Tsin370-0-75-0-note that torque produced by Tcos300- Nx and Ny about the point A is equal to zero-Thus500-40-x2212-T-35-0-75-0orT-540-5-3-0-75-1200N

The tension in the string is:

$600$ N
$80$ N
$1200$ N
$800$ N

For equilibrium about P $\sum τ_{p} = 0$
or
$[(500) 1 + 80 (0.5) - (T s i n 37^{0}) (0.75)] = 0$
(note that torque produced by $T c o s 30^{0}$ , $N_{x}$ and $N_{y}$ about the point A is equal to zero),Thus
$500 + 40 - T (\frac{3}{5}) (0.75) = 0$
or
$T = \frac{540 (5)}{3 (0.75)} = 1200 N$

The tension in the string is:600 N80 N1200 N800 N

For equilibrium about P ∑τp=0or[(500)1+80(0.5)−(Tsin370)(0.75)]=0(note that torque produced by Tcos300, Nx and Ny about the point A is equal to zero),Thus500+40−T(35)(0.75)=0orT=540(5)3(0.75)=1200N

The tension in the string is:

$600$ N
$80$ N
$1200$ N
$800$ N

For equilibrium about P $\sum τ_{p} = 0$
or
$[(500) 1 + 80 (0.5) - (T s i n 37^{0}) (0.75)] = 0$
(note that torque produced by $T c o s 30^{0}$ , $N_{x}$ and $N_{y}$ about the point A is equal to zero),Thus
$500 + 40 - T (\frac{3}{5}) (0.75) = 0$
or
$T = \frac{540 (5)}{3 (0.75)} = 1200 N$