# Some Newton's Laws questions

## Homework Statement

1. A man stands on a scale in an elevator that has an upward acceleration a. The scale reads 960 N. When he picks up a 20kg bos, the scale reads 1200 N. Find the mass of the man, his weight, and the acceleration a.

2. A horizontal force of 100 N pushes a 12kg block up a frictionless incline that makes an angle of 25degree with the horizontal. (a) What i s the normal force that the incline exerts on the block ? (b) what is the acceleration of the block ?

3. A person in an elevator is holding a 10kg block by a cord rated to withstand a tension of 150 N. When the elvator starts up, the cord breaks. What was the minimum acceleration of the elevator ?

4. A 60kg painter stands on a 15kg platform. The platform is attached to a rope that passes through an overhead pulley. (a) To accelerate herself and the platform at 0.8 m/s^2, with what force must she pull on the rope ? (b) what force is she exerting on the rope when her speed reaches 1 m/s and she pulls the rope such that the platform has a constant speed ??

F=ma

## The Attempt at a Solution

1. I did Newton's 2nd law for the man and the man and the box then subtract the system of the man and box by the man to get a. However, I got it wrong.

2. I think the normal force is equal and opposite to the weight so is it -12 * 9.81 ?? For part b, should a = (Fcos25- mgsin25) / m ???

3. I use F=ma. and substitute on the left side as T-mg=ma but I get the wrong answer.

4. I combined 2 mass as one and treat it as one object. Then I use F=ma, left side is F- 75g= ma. I got the wrong answer.

I really need to do these corrections because quarter is ending in a week and I have so much to do this weekend. Thanks a lot.

1. The question seems to be asking for the acceleration of the elevator. Did you account for acceleration due to gravity in addition to the acceleration of the elevator? In other words,
(acceleration of the system) = (acceleration of the elevator) + (acceleration due to gravity)

If you make adjustments for this, you should get the right answer.

2. The normal force will act at a right angle to the surface of the incline. This should have components of both gravity and the force applied to the block.

Part B sounds correct.

3. Your method sounds correct. What did you get for an answer?

4. This is essentially the same problem as #3, so your approach sounds right. Can you show what numbers you substituted and what your answer was?

1. How do I find the acceleration of the elevator ? I am really lost. Can you walk me through the problem ??

2. Can you explain more about what components involve here and how I can find the normal force ?

3. I fix the answered. Thanks

4. SO I have F-mg= ma leading to F= m(a+g) leading to F= (9.81/ 0.8). What's wrong ? Can I have a step by step suggestion here ?

I need help a lot. I need to finish this soon.

1. You seemed to start with the right approach. From what you wrote initially, I have the following equations:

960 N = mman * (aelevator + agravity)

1200 N = (mman + mbox) * (aelevator + agravity)

From these two equations, you have two unknowns - mman and aelevator. Solve one equation for one unknown in terms of the other, then substitute into the other equation.

2. Think about how you got the acceleration up the incline: a = (F cos 25 - mg sin 25)/m. This equation is correct. The normal force will work 90º to this, so the equation will look somewhat similar.

If you haven't already, draw a free body diagram of the system with the forces (gravity and the applied force). The force acting on the block up the incline acts parallel to the incline, while the normal force will act perpendicular to the incline.

4. Using F = m (a + g) is correct. I'm not sure how you went from that to F = (9.81/0.8). Check what you're substituting and where.

I got number 1. Thanks for that.

2. How will the information of perpendicular help me ?

4. I will check number 4 to see where did I do wrong. However, my teacher said that F is not in the system. Thus, we cannot use F for Newton's 2nd law. Is that true ??

With part (b) how can I do it ??

2. How will the information of perpendicular help me ?

A normal force is, by definition, the force perpendicular to the surface. Since you've already found the components of the forces acting parallel to the surface, it is fairly easy to find the components perpendicular - just be careful of the direction of the forces (positive or negative).

4. I will check number 4 to see where did I do wrong. However, my teacher said that F is not in the system. Thus, we cannot use F for Newton's 2nd law. Is that true ??

With part (b) how can I do it ??

I'm not sure I understand what your teacher is saying. You are trying to find the Force, so that is an unknown. You do, however, know mass and acceleration, so F = ma is the most logical equation to use. Again, check your substitution into your formula of F = m (a + g). If you do this correctly, you will get the correct answer.

Part (b) is actually very easy - you can use the same equation as for part (a), but consider what the acceleration of the platform will be if you are maintaining a constant speed.

4. I did exactly like that: F= m(g+a)= 75(9.81 + 0.8) but I got the wrong answer which is around 796 N. Any idea what's going on ??

Another question: A 10kg object on a frictionless table is subjected to 2 horizontal forces, F1 and F2 with magnitude F1= 20 N and F2= 30N. F2 acts on the object with an 30 degree angle and F1 acts vertically down ( x component =0). Find a F3 so that the object is in static equilibrium.

So I set F1+F2+F3= 0 but go the wrong answer. Anything wrong here ??

Another question: A 10kg object on a frictionless table is subjected to 2 horizontal forces, F1 and F2 with magnitude F1= 20 N and F2= 30N. F2 acts on the object with an 30 degree angle and F1 acts vertically down ( x component =0). Find a F3 so that the object is in static equilibrium.

So I set F1+F2+F3= 0 but go the wrong answer. Anything wrong here ??

make sure to break the vectors F2 doesnt act on the object in 1D it acts on it with a 30 deg angle and F1 acts on it in 1D.

I got -25.98 i + 5 j. Can anyone please check it for me ??? I really need it now.

I still don't know how to do the Normal force one. I drew an FBD and I got the accleration right. I can see that normal force is equal and opposite of x component of gravity. Is that true ?? Or what other components do I have to add in ??

Also the one about the platform. Can anyone check it for me ? 4. I did: F= m(g+a)= 75(9.81 + 0.8) but I got the wrong answer which is around 796 N. Any idea what's going on ??

F1 is acting on the object straight down so it affects the j vector of F2. then set the net force of the system to be zero and find F3. make sure you're doing your trig correctly

I am sure that I am doing everything right.

I have: 0= -20 j +(30cos30 i + 30sin 30 j) + F3. and I came up with -25.98 i + 5 j. Am I correct ?

I got -25.98 i + 5 j. Can anyone please check it for me ??? I really need it now.

I still don't know how to do the Normal force one. I drew an FBD and I got the accleration right. I can see that normal force is equal and opposite of x component of gravity. Is that true ?? Or what other components do I have to add in ??

Also the one about the platform. Can anyone check it for me ? 4. I did: F= m(g+a)= 75(9.81 + 0.8) but I got the wrong answer which is around 796 N. Any idea what's going on ??

How are you checking your answers? If you are entering into a computer system that checks for exact answers, check that you are using the right value for gravity (e.g., if g = 9.8 your answer would be 795 N). The other possibility is that the system is moving downward, while the above calculations assumed upwards.

For the Normal Force, you should have two forces to add - a component from gravity and a component from the applied force. You already calculated the parallel components of these forces for the acceleration up the incline - adding the other components will give you the normal force.

I am sure that I am doing everything right.

I have: 0= -20 j +(30cos30 i + 30sin 30 j) + F3. and I came up with -25.98 i + 5 j. Am I correct ?

Just to clarify, are these forces being applied in 2 dimensions? If so, are these dimensions a single horizontal direction and vertically? (in other words, is there a "Z" component, or just "X, Y"?)

If there is a vertical components, you may need to account for forces due to gravity and a reactionary force from the table. The way the question was phrased, there could be a number of correct answers.

THere are only 2 dimensions.

I know it was wrong because my teacher marked it wrong. I am doing corrections for it.

THere are only 2 dimensions.

I know it was wrong because my teacher marked it wrong. I am doing corrections for it.

Unless I'm missing something, I don't see anything wrong with your answer for #4. Unless, of course the motion of the system is downward instead of upward. Sorry I can't help.

For the other problem, consider the possibility that your teacher is allowing the frictionless table to exert a vertical reactionary force. You may only need a force to counteract the horizontal components.

Can you help me with the normal force problem. I am still clueless about how can I find it.

Can you help me with the normal force problem. I am still clueless about how can I find it.

For finding the acceleration parallel to the incline, you got the equation

a = (Fcos25- mgsin25) / m

which is the same as

Fparallel to incline = Fcos 25 - mg sin 25

So to get here, you had to multiply the applied force by the cosine of 25º to get the component that is parallel to the incline. You also had to multiply the force from gravity by the sine of 25º to get the component that is parallel to the incline. When you look at a diagram of the block, you can see that these forces are working in opposite directions (one up the incline, the other down it), so you subtracted one from the other.

Now, when you look at the normal force, you are looking at the other components of these two forces. To get the component of the applied force in this direction, rather than multiplying by the cosine of 25º, you multiply it by the sine of 25º. Does this make sense? The same idea will apply to the force from gravity.

So basically normal force= Fsin25-mgcos 25 ??

Ok. So I got it.

How about the question about the painter and the platform and the question about the object with 2 forces acting upon ( 1 force acts on with a 30 degree angle)

So basically normal force= Fsin25-mgcos 25 ??

Almost. Think about the direction the forces are acting - if they are both in the same direction, add them together.

How about the question about the painter and the platform and the question about the object with 2 forces acting upon ( 1 force acts on with a 30 degree angle)

I honestly have no idea why your answer for the painter and platform problem was not correct.

I'm still a bit confused on the last problem:
A 10kg object on a frictionless table is subjected to 2 horizontal forces, F1 and F2 with magnitude F1= 20 N and F2= 30N. F2 acts on the object with an 30 degree angle and F1 acts vertically down ( x component =0). Find a F3 so that the object is in static equilibrium.

Are F1 and F2 horizontal forces, or is F1 vertical?

The reason I ask is this: if the objects are on a flat table, any force that acts downward in a vertical direction can be ignored, as the table will provide the necessary reactionary force (unless you're calculating what force the table will exert on the object, of course). The only force needed to maintain static equilibrium would be horizontal, equal and opposite to the horizontal component of F2.

F1 is horizontal but pointing down the Y axe while F2 acts on with a 30 degree angle.

So imagine it's like you pull the object both up to right and down at the same time. Get it ?

F1 is horizontal but pointing down the Y axe while F2 acts on with a 30 degree angle.

So imagine it's like you pull the object both up to right and down at the same time. Get it ?

I think I finally got it, thanks for the clarification.

Going by the numbers you posted earlier:
25.98i + 5j

sounds correct, assuming the 30 degree angle is between the force vector and the x-axis.

Does your teacher want the answer in this format, or in a force @ angle format?

is that 25.98 or -25.98 ?

How about the painter problem ??

is that 25.98 or -25.98 ?

How about the painter problem ??

Sorry, I went through it too quickly. -25.98 should be correct for the x-component of the F3 force vector.

Thanks a lot.

How about the painter problem ?