A string of negligible mass going over a clam | vedclass

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(D) The forces acting on the pulley are:$1$. The tension $T$ in the horizontal part of the string,where $T = Mg$.$2$. The tension $T$ in the vertical

Vedclass · Accepted Answer

$\sqrt {{{(M + m)}^2} + {M^2}} g$

Vedclass · Answer

(D) The forces acting on the pulley are:$1$. The tension $T$ in the horizontal part of the string,where $T = Mg$.$2$. The tension $T$ in the vertical part of the string,where $T = Mg$.$3$. The weight of the pulley,$mg$,acting downwards.$4$. The force exerted by the clamp on the pulley,$F_{pc}$.For the pulley to be in equilibrium,the vector sum of all forces must be zero. The forces acting on the pulley are the horizontal tension $T$ to the left,the vertical tension $T$ downwards,and the weight $mg$ downwards.Total horizontal force $F_x = T = Mg$.Total vertical force $F_y = T + mg = Mg + mg = (M + m)g$.The magnitude of the force exerted by the clamp on the pulley is the resultant of these two perpendicular forces:$F_{pc} = \sqrt{F_x^2 + F_y^2}$$F_{pc} = \sqrt{(Mg)^2 + ((M + m)g)^2}$$F_{pc} = \sqrt{M^2 + (M + m)^2} g$

$A$ string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by

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