What Forces Keep a Laptop on a Car Roof at High Speeds?

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Homework Help Overview

The discussion revolves around the forces acting on a laptop placed on the roof of a car while it is in motion. The scenario involves analyzing the forces during deceleration and the effects of friction, with specific parameters such as the mass of the laptop, the applied forces, and the coefficient of static friction provided.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the free body diagram of the laptop, questioning the accuracy of the forces depicted. There is exploration of the maximum acceleration that can be applied without causing the laptop to slide off, with attempts to derive equations based on forces in both x and y directions.

Discussion Status

There is ongoing clarification regarding the normal force and its relationship to other forces acting on the laptop. Some participants express confusion about the calculations and seek confirmation of their reasoning, while others provide insights into the direction of frictional forces and their role in preventing sliding.

Contextual Notes

Participants are navigating through various interpretations of the forces involved, particularly the normal force and friction, while adhering to the constraints of the problem as posed in a homework context.

riseofphoenix
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<< Original post restored by Moderators after being deleted by riseofphoenix >>

While you are packing up to drive to school, you forget your laptop on the roof of your car. As you are driving west towards school at v = 45 mi/hr, you suddenly remember the laptop, roll down your window, and put your hand on the laptop exerting a force of 50 N downward and 20 N horizontally toward the rear of the car on the laptop, and hit the brakes. The coefficient of static friction of between the laptop and car roof is 0.4. The mass of the car is 1200 kg and the mass of the laptop is 4 kg.

(a) Draw the free body diagram of the laptop. Label ALL forces acting on it. Label your coordinate axes defining your directions.

I know that:
v = 20.1168 m/s
F1 (downward) = 50 N
F2 (horizontal) = 20 N
μs = 0.4
fs = 0.4*n
mass of the laptop = 4 kg

Is this how the free-body diagram should look like?
IMAGE: http://i11.photobucket.com/albums/a179/slourg/freebodydiagram_zps7eb1828b.jpg

(b) What is the maximum acceleration (magnitude AND direction) that can be applied to the car so that the laptop does not end up in several pieces on the pavement?

Forces in the x-direction:
20
fs

ƩFx = ma
20 + fs = (4)a
20 + 0.4n = (4)a​

** Solve for normal force, n, from ƩFy = 0 **​
20 + 0.4(-10.8) = (4)a
20 - 43.2 = (4)a
-23.2 = (4)a
-23.2/4 = a
-5.8 m/s2 = a​

QUESTION: Is that the answer? a = -5.8 m/s2? If not, can you show me how to get the answer?

Forces in the y-direction:
n
50
-39.2

ƩFy = 0
n + 50 + (-39.2) = 0
n + 50 - 39.2 = 0
n + 10.8 = 0
n = -10.8​
 
Last edited by a moderator:
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riseofphoenix said:
While you are packing up to drive to school, you forget your laptop on the roof of your car. As you are driving west towards school at v = 45 mi/hr, you suddenly remember the laptop, roll down your window, and put your hand on the laptop exerting a force of 50 N downward and 20 N horizontally toward the rear of the car on the laptop, and hit the brakes. The coefficient of static friction of between the laptop and car roof is 0.4. The mass of the car is 1200 kg and the mass of the laptop is 4 kg.

(a) Draw the free body diagram of the laptop. Label ALL forces acting on it. Label your coordinate axes defining your directions.

I know that:
v = 20.1168 m/s
F1 (downward) = 50 N
F2 (horizontal) = 20 N
μs = 0.4
fs = 0.4*n
mass of the laptop = 4 kg

Is this how the free-body diagram should look like?
IMAGE: http://i11.photobucket.com/albums/a179/slourg/freebodydiagram_zps7eb1828b.jpg

(b) What is the maximum acceleration (magnitude AND direction) that can be applied to the car so that the laptop does not end up in several pieces on the pavement?

Forces in the x-direction:
20
fs

ƩFx = ma
20 + fs = (4)a
20 + 0.4n = (4)a​

** Solve for normal force, n, from ƩFy = 0 **​
20 + 0.4(-10.8) = (4)a
20 - 43.2 = (4)a
-23.2 = (4)a
-23.2/4 = a
-5.8 m/s2 = a​

QUESTION: Is that the answer? a = -5.8 m/s2? If not, can you show me how to get the answer?

Forces in the y-direction:
n
50
-39.2

ƩFy = 0
n + 50 + (-39.2) = 0
n + 50 - 39.2 = 0
n + 10.8 = 0
n = -10.8​
Depending upon the amount of deceleration due to braking, the force of friction may be towards the rear of the vehicle. Under maximum braking, the force of friction will be towards the rear of the vehicle.

The normal force is in opposition to both the weight and the applied force of 50 Newtons.
 
Last edited by a moderator:
<<Original post restored by moderators after being deleted by riseofphoenix>>

Wait...what do you mean?? D=

Is my answer right?

Omg...please don't take offense to thins but I have no idea where you're getting at... :(
Everytime I post a thread like this, I always get extremely vague/abstract answers.
Can you please clarify?

Are you saying normal force DOESN'T equal -50 + 39.2?
 
Last edited by a moderator:
riseofphoenix said:
Wait...what do you mean?? D=

Is my answer right?
No.
Omg...please don't take offense to thins but I have no idea where you're getting at... :(
Everytime I post a thread like this, I always get extremely vague/abstract answers.
Can you please clarify?

Are you saying normal force DOESN'T equal -50 + 39.2?
Yes, that's what I'm saying.

What are the vertical forces acting on the laptop?
 
<<Original post restored by moderators after being deleted by riseofphoenix>>

So what does N equal then?

n + (-50) + (-39.2) = 0
n = 89.2??
 
Last edited by a moderator:
riseofphoenix said:
So what does N equal then?

n + (-50) + (-39.2) = 0
n = 89.2??
Yes.
 
The main force keeping the laptop from sliding (forward) as the car brakes is the force of friction between the roof & the laptop. To keep the laptop from sliding forward, in what direction must the force of friction act on the laptop?

That 20 N component merely aids in keeping the laptop from sliding forward.
 
<<Original post restored by moderators after being deleted by riseofphoenix>>

So it should be...

20 + 0.4(89.2) = (4)a
20 + 35.68 = (4)a
55.68 = (4)a
13.92 m/s2= a

Is that the answer?
 
Last edited by a moderator:
riseofphoenix said:
So it should be...

20 + 0.4(89.2) = (4)a
20 + 35.68 = (4)a
55.68 = (4)a
13.92 m/s2= a

Is that the answer?
The method looks good.

Let's see ... How could we check the answer?
 
  • #10
<<Original post restored by moderators after being deleted by riseofphoenix>>

East! since the car is already moving west.
Right?

So is the direction East?
 
Last edited by a moderator:
  • #11
riseofphoenix said:
East! since the car is already moving west.
Right?

So is the direction East?
If the car's moving west, then yes, east. Also, if the car's moving west then the 20 Newton component is also east.
 
  • #12
From the PF rules:

Neither member accounts nor a member's posts will be deleted on demand. It is up to the discretion of the forum owners and admins. Posts are for everyone's benefit and should be thought of as permanent.

This thread will not be deleted.

Zz.
 

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