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Help with some problems involving forces.

  • Thread starter vdfortd
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  • #1
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Homework Statement


statics1.jpg


The 2800lb car is moving at constant speed on a road with the slope shown. The aerodynamics forces on the car are the drag D=280 lbs, which is parallel to the road, and the lift L = 112 lb, which is perpendicular to the road. Determine the magnitudes of the total normal and firction forces exerted on the car by the road.

Homework Equations


Newton's Laws
[tex]\Sigma[/tex]F=0

The Attempt at a Solution



I made the xy plane as if the slope of the car is the x-axis, and the direction of the Lift force is technically the positive y direction. The car is traveling up the incline, which I made the positive X-axis. So the plane is slanted a bit.

When doing my forces, I figured that you have the Drift for (D) going in the negtive x direction, the lift force (L) going in the positive Y direction, you have the Normal force which is perpendicular to the car also going in the positive Y direction, and then you have the mg force that the car exerts in both the negative X and negative Y direction (since the XY plane is slanted).

So when I add them up:

[tex]\Sigma[/tex]Fx= -FD-FmgSin(15)=0
[tex]\Sigma[/tex]Fy= FN+FL-Fmgcos(15)=0

Plug in what you are given and it becomes:

[tex]\Sigma[/tex]Fx= -270-F2800Sin(15)=0

[tex]\Sigma[/tex]Fy= FN+120-2800cos(15)=0

Since I have the Normal Force as the only variable in the [tex]\Sigma[/tex]Fy, I solve for it and get:

N+120-2704=0
N=2704-120
N=2584 lb

I didn't know what to do find the friction forces, so I simplified the [tex]\Sigma[/tex]Fx and it ended up being -270-2800sin(15)= -995. The answer in this case for friction is positive 995. I don't think I did this problem right and just lucked into these answers. Can someone please clarify/explain this to me?

Homework Statement


statics2.jpg


The person wants to cause the 200lb crate to start sliding toward the right. To acheive this, the horizontal component o fthe force exerted on the crate by the rope must equal .35 times the normal force exerted on the crate by the floor. In figure a, the person pulls on the rope in teh direction shown. In figure b, the person attaches the rope to a support as shown and pulls upward on the rope. What is the magnitude of the force he must exert on the rope in each case?

Homework Equations


Newton's Laws
[tex]\Sigma[/tex]F=0


The Attempt at a Solution



I can't start this problem because I am having trouble with the forces.

For (a)

I know theres a tension force on the rope acting in the positive x and y direction.
There is the force of gravity (mg) acting in the negative y direction.
There is a normal force acting perpendicular to the crate, in the positive why direction.

What i'm confused about is the friction force. It says it's .35 times the normal force acting on the crate, so I guess you can say it's .35(N), but does it just act solely in the negative X direction? and is .35N the right way to calculate it?

For (b)

The man is pulling the rope up from the middle. I have the same question about the friction force here as in part (a), and I was wondering if the wrote now as two tension forces?

The way I see it, you have the Normal force, and force of Gravity as in the first one. Now there is a horizontal tension force from the rope going from the crate to the man, and another tension force going from the man to the ground where the rope is tied down, and this one acts in the negative X and positve Y directions. But now isn't there a force for the man pulling the rope straight up from where he is standing? What would that be? I'm pretty sure I can solve this problem. These forces are just confusing the heck out of me.

Thanks for any and all explanations you guys can give me.
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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Homework Statement


statics1.jpg


The 2800lb car is moving at constant speed on a road with the slope shown. The aerodynamics forces on the car are the drag D=280 lbs, which is parallel to the road, and the lift L = 112 lb, which is perpendicular to the road. Determine the magnitudes of the total normal and firction forces exerted on the car by the road.

Homework Equations


Newton's Laws
[tex]\Sigma[/tex]F=0

The Attempt at a Solution



I made the xy plane as if the slope of the car is the x-axis, and the direction of the Lift force is technically the positive y direction. The car is traveling up the incline, which I made the positive X-axis. So the plane is slanted a bit.

When doing my forces, I figured that you have the Drift for (D) going in the negtive x direction, the lift force (L) going in the positive Y direction, you have the Normal force which is perpendicular to the car also going in the positive Y direction, and then you have the mg force that the car exerts in both the negative X and negative Y direction (since the XY plane is slanted).

So when I add them up:

[tex]\Sigma[/tex]Fx= -FD-FmgSin(15)=0
[tex]\Sigma[/tex]Fy= FN+FL-Fmgcos(15)=0

Plug in what you are given and it becomes:

[tex]\Sigma[/tex]Fx= -270-F2800Sin(15)=0

[tex]\Sigma[/tex]Fy= FN+120-2800cos(15)=0

Since I have the Normal Force as the only variable in the [tex]\Sigma[/tex]Fy, I solve for it and get:

N+120-2704=0
N=2704-120
N=2584 lb

I didn't know what to do find the friction forces, so I simplified the [tex]\Sigma[/tex]Fx and it ended up being -270-2800sin(15)= -995. The answer in this case for friction is positive 995. I don't think I did this problem right and just lucked into these answers. Can someone please clarify/explain this to me?
For part 1, correct your typo; you had L=112 lb, but you used L=120lb. Otherwise, you did well, but you forgot the friction force in your Sum of Fx equation (which you later aded in). The friction force is the force betweeen the car's tires (front tires if Front Wheel Drive) and the road. It is this friction force that drives the car forward in the positive x direction.
 
  • #3
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So when I draw the free body diagram, the friction force is an arrow parallel to slope, and in the direction of the positive X-axis(the car going UP the incline)? I thought that friction forces act in the opposite direction the object is moving?
 
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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So when I draw the free body diagram, the friction force is an arrow parallel to slope, and in the direction of the positive X-axis(the car going UP the incline)?
Yes, when you draw the FBD of the car, the (static) friction force acts on the car in a direction up the plane. By Newton's 3rd law, the friction force of the car on the ground acts in the opposite direction, down the plane.
I thought that friction forces act in the opposite direction the object is moving?
Yes, correct, the tire is rotating clockwise, pushing back (leftward) on the road down the incline. The road pushes on the car up the incline, rightward. It's like walking, you push back on the ground with your feet, and the friction force of the ground pushes you ahead.
 
  • #5
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,136
476
Part 2:
For (a)

I know theres a tension force on the rope acting in the positive x and y direction.
There is the force of gravity (mg) acting in the negative y direction.
There is a normal force acting perpendicular to the crate, in the positive why direction.
yes
What i'm confused about is the friction force. It says it's .35 times the normal force acting on the crate, so I guess you can say it's .35(N),
yes, F_friction max = (mu)N
but does it just act solely in the negative X direction?
yes, the friction force acting on an object always opposes its direction of motion, or tendency of direction of motion, relative to the surface the object it is in contact with.
and is .35N the right way to calculate it?
yes
For (b)

The man is pulling the rope up from the middle. I have the same question about the friction force here as in part (a),
yes, same
and I was wondering if the wrote now as two tension forces?
??
The way I see it, you have the Normal force, and force of Gravity as in the first one. Now there is a horizontal tension force from the rope going from the crate to the man, and another tension force going from the man to the ground where the rope is tied down, and this one acts in the negative X and positve Y directions.
Yup.
But now isn't there a force for the man pulling the rope straight up from where he is standing?
yes
What would that be? I'm pretty sure I can solve this problem. These forces are just confusing the heck out of me.
First look at the horizontal tension required to move the box, by examining the forces on the box. Once you know that tension, you can calculate the tension in the diagonal rope and the upward force of the man on that rope, using Newton 1 at that 'joint' where the 2 ropes and the man's hands meet.
 

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