How Do You Apply Newton's Laws to Calculate Forces on a Lawn Mower?

  • Thread starter Thread starter kgood5885
  • Start date Start date
  • Tags Tags
    Laws Newton's laws
Click For Summary
SUMMARY

This discussion focuses on applying Newton's Laws to calculate forces acting on a 13.0 kg lawn mower being pushed at a constant speed with a force of 84.0 N at an angle of 49.0° to the horizontal. The horizontal retarding force is calculated as 38.22 N, derived from the normal force of 127.4 N and a coefficient of friction of 0.30. To accelerate the mower from rest to 1.3 m/s in 2.0 seconds, users must apply a net force greater than the retarding force, necessitating the use of Newton's second law for further calculations.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Basic trigonometry for resolving forces
  • Knowledge of friction coefficients
  • Ability to apply kinematic equations
NEXT STEPS
  • Learn how to apply Newton's second law in various scenarios
  • Study the concept of friction and how to calculate it in different contexts
  • Explore kinematic equations for calculating acceleration and force
  • Investigate the effects of angles on force components in physics problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the practical applications of Newton's Laws in real-world scenarios, particularly in mechanics involving forces and motion.

kgood5885
Messages
2
Reaction score
0
A person pushes a 13.0 kg lawn mower at constant speed with a force of 84.0 N directed along the handle, which is at an angle of = 49.0° to the horizontal.

(a) Calculate the horizontal retarding force on the mower

(b) Calculate the normal force exerted vertically upward on the mower by the ground.

(c) Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.0 seconds (assuming the same retarding force).

Here's what I have so far...

a) FPx = 84 cos 49 = 55.11 N
FPy = 84 sin 49 = 63.396 N
∑ F = ma
N – mg = ma
N = mg
= 13.0(9.8)
= 127.4 N

b) Ffr = (coefficient of friction)FN
= (0.30)(127.4)
= 38.22 N
c) I have no idea where to even start for this question!

Please help!
 
Physics news on Phys.org
kgood5885 said:
Here's what I have so far...

a) FPx = 84 cos 49 = 55.11 N
FPy = 84 sin 49 = 63.396 N
∑ F = ma
N – mg = ma
N = mg
= 13.0(9.8)
= 127.4 N
All you need to answer this part is to realize that the net horizontal force must be zero. The retarding force must balance the horizontal component of the applied force.

b) Ffr = (coefficient of friction)FN
= (0.30)(127.4)
= 38.22 N
For this part, all you need to find is the normal force. (Where did you get the coefficient of friction?) This time use the fact that the net vertical force must be zero. (Hint: There are three vertical forces.)
c) I have no idea where to even start for this question!
To get an acceleration, the net horizontal force must be greater than zero. Apply Newton's 2nd law.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
6
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
21K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
10K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 3 ·
Replies
3
Views
4K