What are the two forces acting on the ping pong ball and in what direction are they applied?

Because the spin of the ball is not, in such a case, aligned perfectly with the table. If the ball were spinning so that the axis of rotation were exactly perpendicular to the table, then the ball would not move anywhere.

Imagine the tire of a car for example. The axis of rotation of the tire is parallel with the ground, so the outer surface of the tire pushes against the ground.

Same thing with the ball: if the axis of rotation is not perfectly perpendicular to the surface, then the ball will push on the table like the wheel does.

However, the axis of rotation of the ball is nowhere near parallel with the table, so it only pushes a tiny bit.

Spin the ball with an exaggeratedly parallel axis and it will tend to roll more than spin.

EDIT: In the case of the video, the axis of rotation is parallel with the table. The ball will tend to go in the opposite direction of which the bottom surface of the ball pushes, just like the tires on a car.

One of my favorite sports to play is ping pong. I've always had a ping pong table in my basement and play with my brothers and with friends pretty frequently. I've always been amazed at the skill of Olympic table tennis players. If you've never seen Olympic table tennis, It's amazing how fast the little ball is hit back and forth between the two players. Sometimes it goes so fast that you can barely see it. The physics behind the game can explain why these Olympic athletes are so skilled at the game. In serving the ball, a player must throw the ball out of his hand at least 6 inches in the air and then hit the ball so that it bounces once on his/her side and once on his opponent's side. According to Newton's second law of motion, the greater the force that is applied to an object, the greater the object will accelerate. Olympic table tennis players hit the ball with great force, causing it to accelerate so rapidly that the ball can hardly be seen. Although they can serve the ball with such great force, their opponent is still able to return the ball without moving his paddle very much at all. The opponent does not need to worry very much about applying a great force to the ball as it comes towards him/her because of Newton's third law. An action force always has an equal and opposite reaction force. As the ball's inertia causes it to travel towards the opponent, it applies a force to the paddle, and the paddle applies an equal and opposite reaction force, causing the ball to accelerate in the opposite direction. Therefore, the force causing the ball to accelerate back towards the player who served it is caused mostly by the inertia of the ball. Watch this video below showing the best 10 table tennis rallies of all time. 

Homework Statement: A ping-pong ball has a weight of 0.50 N. While in contact with a paddle the paddle exerts a force of 7.8 N at 29° above the horizontal. Relevant Equations: Determine the net force acting on the ball at this instant.

w= 0.50N, Fn= 7.8N, angle= 29 Fnet= Fn*fd* costheta = 7.8*0.50*cos(29 = 3.41

is my answer correct

Answers and Replies

Homework Statement:: A ping-pong ball has a weight of 0.50 N. While in contact with a paddle the paddle exerts a force of 7.8 N at 29° above the horizontal.
Relevant Equations:: Determine the net force acting on the ball at this instant. w= 0.50N, Fn= 7.8N, angle= 29 Fnet= Fn*fd* costheta = 7.8*0.50*cos(29 = 3.41

is my answer correct

What's the logic behind your calculation?

vela

Fnet= Fn*fd* costheta = 7.8*0.50*cos(29 = 3.41

is my answer correct

3.41 cows? No, that's not even a force. How could it be correct?

what do I like the steps to get the answer to the question

Homework Statement:: A ping-pong ball has a weight of 0.50 N. While in contact with a paddle the paddle exerts a force of 7.8 N at 29° above the horizontal.
Relevant Equations:: Determine the net force acting on the ball at this instant. w= 0.50N, Fn= 7.8N, angle= 29 Fnet= Fn*fd* costheta = 7.8*0.50*cos(29 = 3.41

is my answer correct

I cannot think of any situation where it would make sense to multiply two forces. Have you drawn a free body diagram? That should always be the first step.

Do you know how to add forces vectorially?

I do not know how to add forces vectorially?
Yes but I am not sure if it's correct.

I do not know how to add forces vectorially?
Yes but I am not sure if it's correct.

Is that yes you have drawn a FBD? If so, please try to post it. You do not have any chance of solving this question if you do not know how to add forces. Have you not been taught? Any notes?

Plenty of resources on the net, just search for "how to add forces" or "how to add vectors".

We have two forces acting on the ball. the force of gravity, there is no component is the x direction and all of the force is in the negative x direction

i will do the components of both using fcosθ for the x- component and fsinθ for the y-component

We have two forces acting on the ball. the force of gravity, there is no component is the x direction and all of the force is in the negative x direction

i will do the components of both using fcosθ for the x- component and fsinθ for the y-component

Here's an approximate answer. We have a force of ##7.8N## and a force of ##0.5N##, which is small compared to the first. The answer should be close to ##7.8N## in a direction not much different from the direction of the larger force. The answer can't be ##3.41N## or anything like that.

For the Fnet I got 7.57N and the angle has 25.7

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