Question

V Base angles of triangular wedge ABC are 37° & 90°. A uniform rope is kept over the wedge as shown in two figures. The maximum length of string which can be hanged vertically is X1 (fig.(i)) & minimum length is x2 X1 (fig.(ii)) in equilibrium. If = 3, find the value X2 of coefficient of friction between rope & wedge-​

Answer 1

The coefficient of friction between rope and wedge is $\frac{1}{2} ,-3$.

Explanation:

$m_{1}g=m_{2}gsin5x+f$

λ1$x_{1}g=$λ$(l-x_{1})g*\frac{3}{5} *$цλ$(l-x_{1})g*\frac{4}{5}$

$5x_{1}=3(l-x_{1})+4u(l-x_{1})$

$m_{1}g=m_{2}gsin30^{0}-f$

λ$x_{2}g=$λ$(l-x_{2})g*\frac{3}{5} -$λ$(l-x_{2})g*\frac{4}{5}$*ц

$5x_{2}=3(l-x_{2})+4u(l-x_{2})$

$5x_{2}=3(l-x_{2})+4(l-x_{2})$

$8x_{2}=3l-4ul+4ux_{2}$

$\frac{x_{2}(8-4u)}{x_{1}(8+4u)}=\frac{l(3-4u) }{l(3+4u)}$

$32u^{2}+80u-18=0$

$2u^{2}+5u-32=0$

$2u(u+3)-(u+3)=0$

$(2u-1)(u+3)=0$

$u=\frac{1}{2} u=-3$.