Solving a Puzzling Homework Problem: Pushing a Lawn Mower

[iRaid]

Homework Statement

Well I don't think my teacher explained this very well so I need some help.. Thank you in advance.

26.) A person pushes a 14.0-kg lawn mower at constant speed with a force of F = 88.0N directed along the handle which is at an angle of 45.0(degrees) to the horizontal.
a) Draw the free-body diagram showing all forces acting on the mower. Calculate b) the horizontal friction force on the mower, then c) the normal force exerted vertically upward on the mower by the ground. d) What force must the person exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force.

Homework Equations

Fx = F cos (theta)
Fy = F sin (theta)
F = ma
a = (vf-vi)/t

The Attempt at a Solution

a) My answer was:

b) Fx = 88 cos 45 = 62.2N

c) I thought it would be the vertical component, but it's not, not sure where to go from that now..

d) Not sure :(


Thanks.

 

[fizzynoob]

Im pretty sure your Fp should be pointing in the third quadrant, you don't push up on a lawn mower. To find the Normal just sum the forces in the Y direction and solve for Fn

 

[iRaid]

The answer is 199 though :|

 

[fizzynoob]

yes the answer is 199.43 N, you need to correct your diagram

[PLAIN]http://img836.imageshack.us/img836/3172/blockl.png

Now sum the forces in the Y direction

Forces in Y = Fn - Fg - Fpsin(t) = 0;

solve for Fn and you will find your answer

 

[rwisz]

I would first like to commend you for seeking help and clarification on this homework problem. It is important to fully understand the concepts and equations involved in order to accurately solve the problem.

To start, let's draw a free-body diagram for the lawn mower. This will help us visualize all the forces acting on the mower.

In this case, we have the applied force of 88N at an angle of 45 degrees, the weight of the lawn mower (mg), the normal force from the ground (N), and the friction force (Ffr) acting in the opposite direction of motion.

Using the equations Fx = F cos(theta) and Fy = F sin(theta), we can determine the horizontal and vertical components of the applied force.

Fx = 88 cos 45 = 62.2N
Fy = 88 sin 45 = 62.2N

Now, for part b, we can use Newton's second law (F=ma) to solve for the friction force. Since we know the mass of the lawn mower (14.0kg) and the acceleration is zero (since it is moving at a constant speed), we can rearrange the equation to solve for the friction force.

F = ma
62.2N = (14.0kg)(0)
Ffr = 0N

This means that the friction force on the lawn mower is zero since it is moving at a constant speed.

Moving on to part c, we can use Newton's third law (for every action there is an equal and opposite reaction) to determine the normal force exerted by the ground on the lawn mower. In this case, the normal force is equal in magnitude to the weight of the lawn mower (mg).

N = mg
N = (14.0kg)(9.8m/s^2)
N = 137.2N

For part d, we can use the equation a = (vf-vi)/t to solve for the acceleration of the lawn mower. We know the final velocity (vf) is 1.5m/s, the initial velocity (vi) is 0m/s, and the time (t) is 2.5 seconds.

a = (1.5m/s - 0m/s)/2.5s
a = 0.6m/s^2

Now, using F=ma, we can solve for the force needed to accelerate