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# Tag Archives: Net force

## Body at incline

1. In first case we have the box with weight, say, 10 Newtons on incline. We calculate two components (perpendicular and parallel)of weight. We calculate normal force and force of friction. The box is at rest.

2. In second case we have ball on incline, made of same material as box from first case. Weight of the ball is the same as in first case, components of weight are the same as in first case. Normal force and friction have the same amount as in the case of box. But,unlike the box, ball is moving – it slides down. How’s that possible? If calculated net force is zero?

** Petar asked**

Answer:

It is not true that the **ball slides, ** but it will roll down.

In the first case, the forces balance each other and there is no motion.

In the second force, the frictional force acting tangentially backwards (up the incline) and the component of weight of the ball acting parallel to the plane and through the centre of the ball constitute a couple and tends to rotate it. Now there is no sliding; it rolls.

*(If any further clarification is required please post as comment to this post)*

Please refer to http://www.real-world-physics-problems.com/rolling-without-slipping.html for detailed treatment of the Physics of rolling without friction.

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## Force – Impulse – Momentum

Hi All! I’m having mega problems with one part of a physics assignment. Looking for any help.

The details are as follows:

The diagram shows the normal force on Christine’s feet vs. time, as recorded by a force plate while she stands still initially (until point B), then jumps off the plate. (This trial is separate from the one in the previous problem. The graph is over-simplified and idealised, compared to reality.) When her feet leave the plate, the normal force is zero.

1)What is the magnitude of the (upward) impulse generated by the normal force of Christine during the time interval of her jump off the plate?

2)What is the magnitude of the downward impulse due to gravity during this interval?

3)What is the net impulse which propels her upwards when she jumps off the plate? (Recall, the net force on her is the normal force minus the force of gravity.)

4)What is her change in speed upwards for this process?

The graph has NORMAL FORCE (N) on the y-axis and TIME (s) on the X axis.

The line is at a constant 550 N until point B (1.75 seconds) at which time it shoot up vertically to 1550 N at a time of 1.95 seconds. It peaks at this time and position then drops down to 0 N at 2.15 seconds.

Thanks in advance for any guidance that can be provided!

… **Sara**

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## Newton’s Third Law of Motion and some doubts

Newton’s Third Law of Motion states that

“Every action has equal and opposite reaction”

In other words,

“Whenever a body exerts a force on another body, the second body exerts an equal and opposite force on the first body”

Osho Garg asks:

“if a person is sitting on a chair both the chair and person exert force against it so we called that ” there is always an equal and opposite reaction ” but for example a elephant of mass 584 kg sit on a chair, the chair will broke. So what we called for this . Please explain.

Answer:

When an elephant “sits” on a chair, it breaks because it cannot give the “equal reaction” and it “yields” to it.

Therefore, here the force is doing “work”.

**More discussions are welcome**