Make sure you consider the different components of the force separately. In other words, find the force in the x direction and the y direction independently because there is no reason for them to be coupled anyway. Look at the forces from the fish and from the fishing pole (mass) I believe. Add these components up

F^2 = F_x^2 + F_y^2

and then you have the total force.

However, to find the direction consider the flat ground (facing the water) to be ø = 0 degrees and use trig (most likely tan(ø)) to find the angle of the direction. Hopefully you'll realize that the angle is greater than 90 degrees.

Yes! You should try that method our for sure. Just make sure you apply newton's third law (correctly) because without it, your method may as well be impossible.

1st question:

Yes, without slipping means there is no friction

2nd question:

If you use your proposed method you would need to set it up in such a way where you would find a force to evaluate the centripetal force.

However, a better method would be to find how many turns the wheel has to make before the person can go that velocity (distance per second).

In other words, compare the circumference and the distance the person travels in a second. Then, use that information to give you the number of turns per second. Convert that to radians per second (units of angular velocity).

The answer to your question is yes! To clarify, however, this would be an effect in the region x<0 and y>0.

No, there most definitely would be a force on the wire when the wire's are perpendicular to one another. Depending how you orient the wires (keeping them perpendicular to each other always though), the magnetic field on the other wire not being held still will be out of the page or into the page. Which is always perpendicular to the current. Additionally, keep in mind, torque (tau = r x F) is always a property the rotation of a wire is dependent on.

Hint: Initially the Force and current will be perpendicular therefore there is a torque.

Like the problem says, you can treat the trampoline as a vertical spring with a mass attached to the end (the child). You can find the spring constant by setting the spring force equal to the force of gravity on the child. From here, you can use the equations you know to find the angular frequency and the period.

To find when the child goes airborne, first consider the net force on the child when it is in contact with the trampoline. You know the position is a simple sine or cosine with time, so you can differentiate twice to find the acceleration. Multiply by the child's mass and you know the net force that must be on the child.

When the child goes airborne, right when they leave contact with the trampoline, the net force on the child is just the force of gravity. In other words, the maximum amplitude the child could attain before going airborne is when the force of gravity equals the maximum of the force you found above. From here, try solving for the amplitude.

You know that the buoyant force is equal to the weight of the fluid displaced. You also know that for the ship the float, its weight must be cancelled out by this buoyant force. The weight of water displaced is given in the problem, so you know the force that the ping-pong balls must provide.

You can find the volume of a ping-pong ball by treating it as a sphere with the given radius, and then you can find the mass of water displaced by one ping-pong ball using the density of seawater given.

m = ρ V

From the mass you can find the weight of water displaced by a single ping-pong ball, and then find the number of ping-pong balls required to raise the ship.

The formula for the magnitude of torque is τ = r F sin θ where θ is the angle between the F and r.

Use trigonometry to figure out which magnitude is greater.

In general, you only find perpetual motion in nature if an energy source is provided. In this case, there is no pump, and hence no supply of energy. If you're wondering where energy might be lost in a fountain like this, consider the possibility of viscosity, or friction in liquids. Also, the water droplets that collide with the basin might not collide elastically, meaning energy would be lost.

This problem is a static equilibrium problem because there is no net force and no net torque -- nothing is moving.

It's really important that you draw a free body diagram for this problem. Draw the plank and all the forces acting on it. Let's think about how many we have:

- Force of gravity on the plank, pointing down from the center of mass of the plank (the center)

- Weight of the diver, pointing down from the end of the plank

- Force from the inner support, pointing up

- Force from the rear support, pointing ???

Which direction does the force from the rear support point? Our instinct might be to say "up", but imagine what would happen if you took away the rear support. Pretend it's not there and it's clear to see that there is going to be a net torque on the plank; i.e. there would be nothing to keep it from tipping forward! So the force from the rear support must point down to balance out the torques. Once you believe this, set the net force and net torque equal to zero and solve for the forces.