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
AI Thread Summary
A person pushes a 13.0 kg lawn mower with a force of 84.0 N at a 49.0° angle, requiring calculations for horizontal retarding force, normal force, and acceleration. The horizontal component of the applied force is 55.11 N, while the normal force is calculated to be 127.4 N, resulting in a frictional force of 38.22 N using a coefficient of friction of 0.30. To accelerate the mower from rest to 1.3 m/s in 2.0 seconds, the net horizontal force must exceed zero, necessitating the application of Newton's second law. The discussion emphasizes balancing forces to determine the required calculations. Understanding these principles is essential for solving the problem accurately.
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top