Question

A 7.30 kg sign hangs from two wires. The first wire is attached to the left end, and pulls 28.0 N directly left. What is the y-component of the force of the second wire?

Answer 1

Approximately
`71.6\; {\rm N}` (assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.)

Step-by-step explanation:

Refer to the diagram attached. Forces on this object are:

• Tension on the left, from the wire on the left end.
• Tension on the right, from the wire on the other end.
• Weight, from the planet.

Assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`, the magnitude of the weight of the sign would be:

`\begin{aligned}(\text{weight}) &= m\, g \\ &= (7.30\; {\rm kg})\, (9.81\; {\rm N\cdot kg^(-1)}) \\ &\approx 71.6\; {\rm N}\end{aligned}`.

Note that weight points downwards (negative) and is entirely in the vertical direction. As a result, the
`y`-component of weight would be equal to
`(-71.6)\; {\rm N}`.

Hence, the
`y`-component of these forces would be:

• 
  `0\; {\rm N}` for the wire on the left end, since this tension is entirely horizontal (entirely in the
  `x`-direction,)
• 
  `(-71.6)\; {\rm N}` for the weight, which points downwards, and
• Not yet found for the tension from the other wire.

Since forces on the object to be balanced, forces need to be balanced in each component. For forces in the
`y`-component to be balanced, forces in the vertical direction need to add up to
`0\; {\rm N}`:

`0\; {\rm N} + (-71.6)\; {\rm N} + (\text{$y$-component of tension on the right}) = 0\; {\rm N}`.

Hence, the
`y`-component of the tension from the wire on the right end would be
`71.6\; {\rm N}`.