Midterm Today- Help calculating friction and forces PLEASE

[Brandiiiii]

Problem:
A man mows his lawn and pushes the mower at a constant speed. The effective coefficient of kinetic friction between the mower and the lawn is 0.40. He pushes on the lawn mower handle at an angle of 45 degrees to the horizontal. (Assume the lawn surface is horizontal.) The mower has a mass of 25 kg. Solve for the normal force acting on the lawn mower, the effective kinetic friction acting on the lawn mower, and the magnitude of the force exerted by the man.


I keep getting stuck in my equations because I don't have enough information. The normal force I have calculated by multiplying the mass (m) by the acceleration due to gravity (professor told us to use 10 instead of 9.8 for simplicity), then subtracting that number from itself multiplied by the sine of 45 degrees. This gave me 426.78 N for the normal force.
I used this value, despite my doubts at its accuracy, to find the kinetic friction. I multiplied the friction coefficient (0.40) by the normal force (426.78 N hopefully) and got the value 170.71 N.
Finally I tried solving for the force exerted by the man and completely confused myself. Physics is an enigma to me. Help would be tremendously appreciated.

 

[Doc Al]

I keep getting stuck in my equations because I don't have enough information. The normal force I have calculated by multiplying the mass (m) by the acceleration due to gravity (professor told us to use 10 instead of 9.8 for simplicity), then subtracting that number from itself multiplied by the sine of 45 degrees. This gave me 426.78 N for the normal force.

I don't understand the logic of what you're doing. Instead, analyze the forces acting on the lawnmower. First: draw a diagram showing all the forces acting on the lawnmower. How many forces act? What are they?

Hint: Label the force that the man exerts as "F".

 

[Brandiiiii]

Thank you so much for your reply! Honestly I don't understand the logic of what I was doing either. I just found the formulas in my notes and tried to make them work.
I drew a free-body diagram and it contained these forces: n (normal force), f (friction), mg (force of gravity), and F (force exerted by the man).
The formula I have to find the normal force is n= mg (cos theta). Is this the correct one to use?
To find the kinetic friction I used the formula f= un where u is the coefficient of kinetic friction. However, if I didn't calculate the normal force correctly this would be inaccurate also.
To calculate the force exerted by the man wouldn't I need a value for acceleration first? I used the following formula to calculate the acceleration: F(cos theta) - f / m = a
and then used the acceleration in the formula F=ma to determine the force.

 

[Doc Al]

I drew a free-body diagram and it contained these forces: n (normal force), f (friction), mg (force of gravity), and F (force exerted by the man).

Excellent!

The formula I have to find the normal force is n= mg (cos theta). Is this the correct one to use?

No. That's for an object on a surface when no extra force is applied. But here we have the force applied by the man, so you'll need to do something else to solve for the normal force.

To find the kinetic friction I used the formula f= un where u is the coefficient of kinetic friction. However, if I didn't calculate the normal force correctly this would be inaccurate also.

That's the correct formula for the kinetic friction.

To calculate the force exerted by the man wouldn't I need a value for acceleration first? I used the following formula to calculate the acceleration: F(cos theta) - f / m = a
and then used the acceleration in the formula F=ma to determine the force.

The 'trick' here is that there is no acceleration: You are told that he pushes the lawnmower at constant speed.

Do this. Set up two equations, one for vertical force components, another for horizontal force components. What must the net force be in each direction? You'll get two equations which you can solve together for your two unknowns: F and N.