Begin by diagraming the forces acting on the mass in the problem:

The mass itself creates a force due to gravity in the downward direction:

For the mass to remain stationary, the forces must be in equilibrium. Therefore the sum of forces in the x and y directions each must be zero:

The x and y tensions can be written in terms of the magnitude of the tension in each cable. Begin with the x direction:

Because the angle is the same for each, the tension in each angle must be equivalent. Use this property when performing the force balance in the y direction:

The free body diagram of the block is given above. This block has three forces acting on it. First, it's weight under the influence of gravity, which is given as . Second, the normal force of the plane, which is given as . Third, the friction force, which acts opposite to its direction of motion and is given as . We choose a coordinate system so that our x-axis aligns with the motion of the block down the plane, and the y-axis aligns with the direction of the normal force. Thus the friction force points in the negative direction of the x-axis, and the normal force is aligned with the positive direction of the y-axis. However, the weight is not along either of these axes, so we resolve the force into its components, along the negative y-axis, and along the positive x-axis.

Now we can use Newton's 2nd law to relate the given forces above. Newton's 2nd law gives us two equations:

and

Because the block is constrained to move along the surface of the inclined plane, there should be no acceleration in the y direction, and so . Also, because the block moves at constant velocity down the plane, Newton's 1st law assures us that there is no acceleration in the x direction as well, therefore . Plugging these accelerations in, we find that and

Summing all the forces in the x-direction gives us

Summing all the forces in the y-direction gives us

Plugging these values into the force equations above gives us the following equations:

Solving for in the second equation gives us . Thus the normal force is equal to the cosine component of the weight. Substituting in for in the first equation will give us the following:

Now we solve the equation for . Adding to each side gives us:

Now we divide each side by to obtain:

The final result is obtained by canceling the factor and using the triginometric identity:

Therefore we arrive at the conclusion that

The normal force is perpendicular to the plane:

First, we need to find .

We can solve for using the trigonometric equation that applies in this instance. We know the length of the side opposite of (5 m) and the length of the side adjacent to (10 m), so we can use the following equation to solve for :

Rearranging to solve this equation for , you get

Substituting in the side lengths of the given triangle, we can solve for .

Note that the normal force is one of the legs of another right triangle. The other leg is the parallel force, and the hypotenuse is the force of gravity.

Using trigonometry, we know that

because , or, in terms of this problem, .

Substituting in the known values into this equation, we can solve for the normal force:

Solving for

Plugging in values

Conservation of momentum gives that the angular momentum of the golf swing is transferred into linear momentum for the ball, or that , so that the velocity of the ball immediately after the swing is .

Using kinematics, we know that , so that substituting gives .

Kinetic friction can be expressed as , so by substituting for in the previous equation, we get

Plugging in known values and solving for gives and

Using

Converting to and plugging in values.

The tensions in both ropes are caused by the bucket being pulled down by gravity, hence the only relevant forces acting in this scenario are those in the y-direction. Since the bucket is being held up against gravity, the upward forces should balance exactly with the downward forces. The forces in the y-direction can be balanced like this:

, where is the tension in rope 1. Thus we find that:

Imagine two men of equal strength pulling on ropes that are connected to a crate. In order to obtain the largest resulting force the two men should pull in the same direction at . Pulling on opposite ends at would result in zero resultant force and anything besides would cause the force to act in an unneeded direction.

As can be seen in the diagram, the force pushing down the ramp is equal to

The force pushing the object into the ramp is

The force into the ramp will be equal to the normal force, thus

The net force pushing the woman down the ramp is

At the moment she starts slipping, the net force is equal to zero

Solving for :

Before we start using equations, we need to determine what forces are acting on the block in this system. The only relevant forces in this situation are gravity and friction. We are given the acceleration of the block, giving us the tools to find the net force.

Using Newton's second law, we can write:

The force of friction is subtracted because it is in the opposite direction of the movement of the block. Substituting in expressions for each variable, we get:

Canceling out mass and rearranging for the coefficient of kinetic friction, we get:

We have values for each variable, allowing us to solve: