Auniform beam of length l and mass m has its lower end pivoted at p on the floor, making ạn angle θ with the floor. a horizontal cable is attached at its upper end b to a point a on a wall. a box of mass m is suspended from a rope that is attached to the beam one-fourth l from its upper end otheexpertta. com 25% part (a) write an expression for the y-component py of the force exerted by the pivot on the beam(b) write an expression for the tension t in the horizontal cable ab 25% part (c) write an expression for the x-component px of the force exerted by the pivot on the beam, in terms of t. 25% part (d) what is the tension in the horizontal cable, in newtons, if the mass of the beam is 49 kg, the length of the beam is 6 m, and the angle is 41°?

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge

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