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## 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?

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. ## Where is the light energy going?

where is the sun’s light enery going once the sun has set?

or

where is the light going when we switch off the torch facing an object. if light gets reflected how long it will do so. and if it gets absorbed, then will it add to the potential energy of the object that is facing the torch?

or Is there any cycle (like hydrocycle)to prove that energy can neither be created nor be destroyed? ## 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 ## Conservation of linear momentum – Problem

Astronauts Mr.X and Mr.Y float in a zero gravity space with no relative velocity to one another. Mr. Y throws a mass of 5 kg towards X with speed 2 m/s. If Mr. X catches it, the change in velocity of X and Y are (choose the correct option)
(a) 0.21 m/s, 0.80 m/s
(b) 0.80 m/s, 0.21m/s
(c) 0.12 m/s, 0.08 m/s
(d) 0.08 m/s, 0.12 m/s

Please provide me the solution part step by step ## Rotational Equilibrium – A Numerical Problem

A father and his son are supposed to carry a load of 500 N. They decided to use a uniform pole that is 4.0 m long and weighs 60 N. The father and his son support the pole at the ends. Where should the load be placed so that the father supports three times as much as his son?

Posted by Daniel

Let’s represent the situation be the folowing diagram Let the force exerted by son be f

Then the force exerted by father = 3f

Therefore the total force exerted =Total weight = 560N = 4f

Which gives f = 140 N, the force exerted by son

3f=420 N, the force exerted by father

Applying the principle of moments about the point where father holds the rod,

420 x 0 – 500 x x – 60 x 2 + 140 x 4 = 0

==> x = 440/500 = 0.88 m from the end where father holds the load

(If there is any problem you find in the solution given, please post as comment to this post.) ## Escaping from gravity

Hello everyone, ever since I heard of the concept of a gravity well I’ve had a great deal of trouble coming to terms with the fact that gravity’s actual pull in infinite ( though admittedly very small at large distances due to the inverse square law ). Is there any way someone could explain to me how it is possible to escape the effects of something with infinite range? ## Is the gravitational force between moon & earth are equal ? Newton’s law of universal gravitation for two bodies. This law governs gravitational forces in the Earth. (Photo credit: Wikipedia)

Is the gravitational force between moon & earth are equal ?

If the question is- “Is the force exerted by earth on moon is equal to the force exerted by moon on earth?” then the answer is “Yes

I hope that more questions will follow.

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