what is the magnitude of the net force on the first wire in (figure 1)?

by Harry Harry
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The magnitude of the net force on a wire is determined by the sum of all forces acting on that wire. This includes gravity, tension from the string, and any other external forces. Figure 1 depicts two wires with different tensions in them. The first wire has a tension of 22N while the second has only 4N. When you find the resultant force for each wire separately, it is easy to see how much more weight there is on one than another (44 N vs 16 N). You may also calculate what this would be when combined into one force: 26N.

What Is The Magnitude Of The Net Force On The First Wire In Figure One?

There are many components to measure in order to determine the magnitude of force on a wire, but we will be focusing on two: weight and tension. The first component is weight which can be found by multiplying mass times gravity. This would mean that 22N multiplied by the gravitational constant g (which equals 980) gives us an answer of 21,840 N/m².

Next, we need to calculate the tension in each wire separately using equation T=W/A where W is what’s known as “tension” or pull from one end of a string while A stands for the surface area between both ends. For example, our second wire has only 16N added due to its smaller surface area with respect to it’s length so this value becomes 0.031 kg*m/s².

If we take the tension of each wire and compare it to what’s known as “weight” or gravitational pull, which is calculated by multiplying mass times gravity (22N multiplied by 980) for an answer of 21,840 N/m² then our first wire has a net force that weighs more than 22N while its counterpart only weights 16N. This means there is a magnitude different in weight between both wires placing greater forces on the first due to their differing values in surface area and weight.

what is the magnitude of the net force on the first wire in figure one? how do you find out what side has less resistance when two wires are stretched taut from opposite ends?

One way to find out which wire has less resistance is by using Coulomb’s law, which says that the side of a taut wire with greater magnitude of force will have more resistive tension. In this case, since our first wire has a net force weighing more than 22N while its counterpart only weighs 16N then it would be safe to say that the first string (the one on the left in Figure I) may experience some resistive tension due to its weight and surface area compared with what seems like negligible amounts for the other string on right. This means there is an advantage for wires on surfaces with greater forces exerted upon them as they are likely stronger because their materials can handle these heavier loads without breaking down or having any other adverse effects.

The magnitude of the net force on the first wire in Figure is approximately 12.0 N, as shown by equation (b) and illustrated in Figure . The four forces acting upon this wire are tension T from the other wires, with a magnitude equal to 18.0N; friction F between the two ends of the wire that has an opposing component 15.0N and a parallel component 0.00N; weight W of 16kg applied at one end which scales down to zero when we consider it as uniform distribution over all length L since L cancels out with denominator 144mm*L/144000mm^-“g” for gravity constant g=32 ft/sec² = 32ft/sec²; and the normal force N, which is perpendicular to each of these three forces.

The magnitude of tension T from other wires 18N can be found by substituting equation (a) into equation (b), resulting in 12N for Figure . This figure also shows that friction F has a net-opposing component 15N, while weight W at one end cancels out with its opposing component of 0.00N because it acts uniformly over all length L as shown by equation(c). The only remaining nonzero vector on this wire is then gravity or the normal force FN of 12N.

The magnitude of the net force on the first wire in Figure is 12 N.

The magnitude of the net force on wire “A” is equal to that of weight “W.”

This is because there are no other forces acting on this first wire. The only act in play at all is gravity, and it has an evenly distributed effect over the entire length of the wire. For example, if one end were held up while another was down, then more gravitational pull would be experienced by that end as opposed to its counterpart. This does not happen with wires because they extend infinitely in both directions from a given point; therefore their weights balance out so long as they remain straight and do not bend or curve away from each other.

If we suppose for a moment that something acted upon Wire “A” to make it bend, then there would be a net force applied by that external agent. If the wire were pulled down and away from Wire “B,” for instance, then the magnitude of Fnet = W + (Mg) on Wire “A” is greater than 12 N.

what are some forces in play between two wires?

What will happen if one wire pulls another towards them? What happens when they push each other away?

What does an electric current do to magnetic fields nearby? Could this cause a change in their strength or directionality?

The following bullet points can also provide discussion topics related to these questions:

How strong are electromagnetic forces relative to gravitational ones as a function of distance?

What are the relative strengths and directions of magnetic fields from different wires that make them repellent or attract one another, if at all?

Are these forces more sensitive to materials properties than gravitational ones (for example, steel versus water)?

How do magnets align with an electric current in a wire loop if they’re not “pushed” by the current itself as it flows through their material volume but rather rely on changing magnetic fields reacting to changes in proximity to currents flowing nearby?

What can be said about this net force caused by external agents acting on Wire “A”? If nothing acted upon Wire “A,” then there would have been no net force applied. This is because without any outside force acting upon Wire “A,” the forces from both wires would have cancelled each other out.

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