- #1

issacnewton

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a) What is the tension force in the wire ?

b) What is the normal force of the wall pushing against the left end of the beam ?

c) What is the upward force supporting the left end of the beam ?

d) How far from the wall should sign be hung so that it is in equilibrium ?

Now following is my attempt at solution. Let ##T## be the tension in the wire and ##\theta = 30 ^{\circ}## be the angle made by the wire with the horizontal. Let ##F## be the upward force supporting the left end of the beam and ##N## be the normal force of the wall pushing against the left end of the beam. Let ##L=4m## be the length of the beam and ##M = 1 kg## be the mass of the beam and ##m=3 kg## be the mass of the cafe sign. Then the condition that the net force on the beam is zero leads to the following equations

$$ N = T\cos(\theta) \cdots\cdots (1)$$

$$ F + T \sin(\theta) = Mg + mg \cdots\cdots (2)$$

Now, the net torque about the right end of the beam is zero. So this leads to the equation

$$ Mg(L/2) +mg(L-x) - FL = 0 \cdots\cdots (3) $$

Also, the net torque about the leftend of the beam is zero. So this leads to the equation

$$ T \sin(\theta) L -Mg(L/2) - mg x = 0 \cdots\cdots (4) $$

Now, it seems that, these four equations are not sufficient to entirely solve this problem with four unknowns ##F,N,T,x##. I just get three equations and 4 unknowns after doing some algebra. So any hints here ?