# Newton's Law Problem (got the first half)

• AznBoi
In summary, a 5kg penguin sits on a 10kg sled and a horizontal force of 45 N is applied to the sled. The penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and snow as well as that between the sled and the penguin is 0.2. To determine the tension in the cord, the penguin's weight and the normal force of the snow on the sled must be taken into account. The tension force of the string is 9.8 N and the acceleration of the sled is 0.3866 m/s^2. The sled's acceleration is due to the forces acting on it such as the
AznBoi
I got the first half of this problem, I'm just stuck on the others. Please take your time to help Thanks a lot!

Problem: A 5kg penguin sits on a 10kg sled. A horizontal force of 45 N is applied to the sled, but the penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and snow as well as that between the sled and the penguin is .2

b) Determine the tension in the cord and c) the acceleration of the sled

if you try to solve for the tension using F=ma, do you could both the weights as the normal force? or do you only count the penguin's weight?

Thanks

Last edited:
AznBoi said:
I got the first half of this problem, I'm just stuck on the others. Please take your time to help Thanks a lot!

Problem: A 5kg penguin sits on a 10kg sled. A horizontal force of 45 N is applied to the sled, but the penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and snow as well as that between the sled and the penguin is .2

b) Determine the tension in the cord and c) the acceleration of the sled

if you try to solve for the tension using F=ma, do you could both the weights as the normal force? or do you only count the penguin's weight?

Thanks
If the penquin holds on tight to the rope, she ain't going nowhere with respect to the ground. Not moving at all with respect to the ground. What does that tell you about the tension force?

tension force includes both objects?? because the sled is the only one being moved? So does that mean you need to include both normal forces when finding the tension of the string?

AznBoi said:
tension force includes both objects?? because the sled is the only one being moved? So does that mean you need to include both normal forces when finding the tension of the string?
No, you've got to take the Free Body Diagram of the bird alone (isolate the bird by drawing a 'cloud' around her only. The cloud cuts thru the cord and just above the sled. Examine the contact and gravity forces acting on her). The tension pulls on her to the left. Since she's not moving w/ respect to the ground, some other force must be pulling on her to the right, equal in magnitude to the tension force. What is that force?

The friction force?? You need to use normal force in the equation. So you would only take the normal force of the penguin??

for the tension force of the string, I got 9.8 N because the friction force equals .2 x 49N. Is that correctly done?

AznBoi said:
for the tension force of the string, I got 9.8 N because the friction force equals .2 x 49N. Is that correctly done?
Yes! Now what do you get for the sled acceleration?

I got .3866 m/s^2 for the sleds acceleration

AznBoi said:
I got .3866 m/s^2 for the sleds acceleration

Sum of all forces sub x = ma sub x

Fg + F sub f + (-Fx) = ma sub x

((.2)(98N+49N))+(9.8N)+(-45N)=15kg(a)

(29.4N)+(9.8N)+(-45N)= (15kg)(a)

-5.8N = 15kg (a)

a= -5.8N/15kg

a=.3866 m/s^2

thanks a lot for helping me btw! I really appreciate it!

AznBoi said:
Sum of all forces sub x = ma sub x

Fg + F sub f + (-Fx) = ma sub x

((.2)(98N+49N))+(9.8N)+(-45N)=15kg(a)

(29.4N)+(9.8N)+(-45N)= (15kg)(a)

-5.8N = 15kg (a)

a= -5.8N/15kg

a=.3866 m/s^2
Almost! You have correctly identified the forces acting on the sled in a FBD of the sled ALONE, and set the sum of those forces equal to ma. But you used 15 for the value of m. What should you have used? And in what direction is the sled accelerating??

So is that correct?

Aren't you suppose to add the two masses together?? The sled is acting away from the rope to the right--->>

Do you use 10 kg because only the sled is accelerating??

Would acceleration be -.58m/s^2?

AznBoi said:
Aren't you suppose to add the two masses together?? The sled is acting away from the rope to the right--->>
Didn't you say this calculation was to find the acceleration of the sled?

AznBoi said:
So is that correct?
No.
When you isolate the sled in a FBD, then the only mass you must use is the mass of the sled, not the mass of the system. The bird's mass should not enter into that equation. Only its normal contact force does (49N acting down on the sled, which, when looking at the normal force of the snow on the sled, translates to the snow to sled normal force of 147N which you correctly identified. Botom line...use m=10 in that equation, not 15.

Oh I get it now, so Would acceleration be -.58m/s^2?

AznBoi said:
Do you use 10 kg because only the sled is accelerating??
No, you use 10 kg because that's the only mass involved in your FBD of the sled. When you do a FBD of the sled, draw a cloud around it. Thje cloud does not include the bird. You must then look at every single contact force that acts on it, plus the gravity force of the sled acting on it. So in your FBD of the sled, you have , in the vertical direction, the weight of the sled acting down, the normal force of the bird on the sled acting down(its weight), and the normal force of the snow on the sled acting up. You solve for the normal force of the snow on the sled by noting that there is no accelerataion in the vertical direction, hence the snow to sled normal force is 147N. Then in the x direction, you have already identified the forces acting on the sled: the 45N applied force, the friction force bird to sled, and the friction force snow to sled. Set it equal to ma, but m is 10! the bird's mass does NOT show up in the FBD. BTW, correct answer is a = 0.58m/s/s.

ok thanks a bunch for your help! I get it completely now

## 1. What are Newton's three laws of motion?

Newton's first law states that an object will remain at rest or in motion with a constant velocity unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The third law states that for every action, there is an equal and opposite reaction.

## 2. How do Newton's laws apply to everyday life?

Newton's laws can be observed in various aspects of everyday life, such as riding a bike (first law), pushing a shopping cart (second law), or launching a rocket (third law). They also play a crucial role in understanding the behavior of objects in motion, such as cars, planes, and satellites.

## 3. What is the difference between mass and weight according to Newton's laws?

Mass is the measure of an object's inertia, while weight is the measure of the force of gravity acting on an object. According to Newton's second law, an object's mass affects its acceleration, while its weight affects the force required to move it.

## 4. Can Newton's laws be broken?

No, Newton's laws are fundamental principles of physics that have been extensively tested and proven to be valid in all situations. However, in extreme conditions, such as near the speed of light or in the quantum realm, they may need to be modified.

## 5. How did Newton's laws contribute to the development of modern science?

Newton's laws revolutionized the way we understand the physical world and laid the foundation for classical mechanics. They also influenced other areas of science, such as astronomy, by providing a mathematical framework to explain the motion of celestial bodies. Newton's laws are still widely used in modern science and engineering to solve various problems and make predictions about the behavior of objects in motion.

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