weight fourmula of ractangular box.

Question

fourmula of weight of rectangular when given hight ,weight .length ,thickness.

Answer 1

Let H=height, W=weight, L=length, B=thickness (breadth), T=constant thickness (of sides, top and bottom of closed box), D=density of box material.

External dimensions: H, L, B. Volume=HLB.

Internal dimensions: H-2T, L-2T, B-2T.

Internal volume=(H-2T)(L-2T)(B-2T)=(H-2T)(LB-2LT-2BT+4T²),

Internal volume=HLB-2HLT-2BHT+4HT²-2LBT+4LT²+4BT²-8T³.

Volume of box material: HLB-(H-2T)(L-2T)(B-2T)=2T(HL+BH+LB)-4T²(H+L+B)+8T³.

Weight W=(2T(HL+BH+LB)-4T²(H+L+B)+8T³)D=2TD(HL+BH+LB-2T(H+L+B)+4T²).

Here's a different approach. Imagine making the box, using glue.

Start with the base and lid each with volume=length×breadth×thickness=LBT. So total volume of base and lid=2LBT.

Next two sides. The height of two of the sides will be reduced by the thickness T of the base and lid. But the length is unaffected.

These two sides are then glued to the base.

Volume of each is (H-2T)LT (height×length×thickness), so total volume of two of the sides is 2(H-2T)LT.

The remaining two sides will be reduced in height by 2T and in breadth by 2T.

Volume of each=(H-2T)(B-2T)T and total volume of these two sides is 2T(H-2T)(B-2T).

These two remaining sides are also glued to the base, and then the lid can be glued to all four sides.

The box has now been constructed so the total volume of material is:

2LBT+2LT(H-2T)+2T(H-2T)(B-2T)=

2LBT+2LHT-4LT²+2T(HB-2HT-2BT+4T²)=

2LBT+2LHT-4LT²+2HBT-4HT²-4BT²+8T³=

2T(HL+BH+LB)-4T²(H+L+B)+8T³, which is the same volume as before.

The weight W is this volume multiplied by the density of the material.