Okay, so I totally don't understand what to do with these three. Could you please show how to solve? I don't need to see the solution, just what steps I need to take for each (every smallest detail please).

1. A box weighing 150 N is at rest on a horizontal floor. The coefficient of static friction is 0.4. What is the smallest force F exerted eastward and upward at an angle of 350 with the horizontal that can start the box in motion?

2. A 450 kg fully loaded elevator is moving downward. The tension in the cable is 3500 N. What is the elevator's acceleration?

3. A 55 kg box slides across the floor by a force of 25N [there was a picture to go along with this, it showed a dotted line going horizontally from the top of the box and making a 45 degree angle].
a.) How large a friction force impedes the motion of the box?
b.) How large is the normal force?

I wish I could help but I hate physics and it would take me forever to find your answer and forever to show you the steps

1. A box weighing 150 N is at rest on a horizontal floor. The coefficient of static friction is 0.4. What is the smallest force F exerted eastward and upward at an angle of 350 with the horizontal that can start the box in motion?

****The force is pulling horizontally, and upward (reducing weight).
normal force=mg - Fcosine350
horizonal force=Fsine350=forcenormal*mu

F sin350=(mg-Fcos350)mu
solve for F
2. A 450 kg fully loaded elevator is moving downward. The tension in the cable is 3500 N. What is the elevator's acceleration?
****Tension=mg-ma
solve for a

3. A 55 kg box slides across the floor by a force of 25N [there was a picture to go along with this, it showed a dotted line going horizontally from the top of the box and making a 45 degree angle].
a.) How large a friction force impedes the motion of the box?
b.) How large is the normal force?
***Work this the same as number one.

Sure! I can help you with each of these problems. Here are the steps you can follow to find the solutions:

1. To find the smallest force F that can start the box in motion, we need to consider the force of friction acting on the box. Here are the steps:

a. Calculate the maximum force of static friction using the equation: static friction = coefficient of static friction * normal force. Here, the normal force is equal to the weight of the box, which is 150 N.

b. Determine the components of the force F in the horizontal and vertical directions. Use trigonometry to find the horizontal and vertical components of the force F using the given angle.

c. Compare the horizontal component of the force F with the maximum force of static friction. If the horizontal component is smaller than the maximum force of static friction, it means the box can start moving. If the horizontal component of the force F is equal to or greater than the maximum force of static friction, the box will not move.

2. To find the elevator's acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object times its acceleration. Here are the steps:

a. Determine the net force acting on the elevator. In this case, the tension in the cable is the force acting on the elevator. Since the elevator is moving downward, the net force will be the tension minus the weight of the elevator, which is equal to mass times acceleration due to gravity.

b. Rearrange Newton's second law equation to solve for acceleration: acceleration = net force / mass.

3. For this problem, we need to determine the friction force and the normal force acting on the box. Here are the steps:

a. Calculate the horizontal component of the applied force by using the given angle of 45 degrees. This can be done using trigonometry.

b. Determine the net force acting on the box in the horizontal direction. The net force is equal to the applied force minus the force of friction. Since the box is moving at a constant velocity (no acceleration), the net force is zero.

c. Solve for the force of friction using the equation: force of friction = applied force. If the force of friction is larger than the maximum force of static friction (which is the same as the applied force), then the box would not be moving.

d. Calculate the normal force acting on the box, which is equal to the weight of the box. The normal force acts perpendicular to the floor.

Remember to double-check the values you use in your calculations and make sure to use the correct units for each quantity. I hope this helps you solve the problems!