Moving About Help (1 Viewer)

[bujolover]

Hi guys,
I have some questions from the Moving About module...

Q1. I need help with understanding why the given answer to this question (see below) is correct. I just don't get it. Shouldn't the vector adjacent to theta (pointing up) be pointing the opposite direction, considering addition of vectors (tail of first vector to head of second one)??? Please explain it to me, and correct me if I'm wrong. :]



Q2. (highlighted one, please)... (I keep getting 2.998 m/s (4s.f.) 89°58' (nearest min), while they get 3.3 m/s East, the bearing of which is essentially the same.)



Q3. Name and describe one safety device in cars that helps protect people from injury. Include the concept of stopping distance in your response. (2 marks)

Thanks in advance!!

P.S. You don't have to answer all the questions at once. Any help is help!

 

[trea99]

Re: Moving About Help ~ Inclined Planes

With the first question;
The only forces acting on the truck would be the weight force, some sort of driving force up the hill (without some force acting up the incline and the in the absence of friction, there would be a net force mgsinθ down the hill but we are told the truck is moving up at a constant speed), and the normal reaction force.
The normal reaction force, N is the vector pointing upward, the weight force is the other vector adjacent to θ pointing down and the small vector opposite θ is the driving force up the hill. There is no "resultant" force vector in this free body diagram because we are told the truck is moving at a constant speed up the hill at the same incline θ which means it is not accelerating (there is no net force acting on it).
We can split the forces acting on the truck to a component parallel to the inclined surface and one that is perpendicular to the incline.
Perpendicular: We have mgcosθ, the component of the weight force perpendicular to the surface and N, the normal reaction force
mgcosθ = N there is no vertical acceleration
Parallel: We have mgsinθ the component of the weight force parallel to the surface and a driving force we'll call Fd
Fd = mgsinθ as well because the truck moves with a constant velocity.
So when we add these vectors, they all "cancel" so we get a closed loop. There is no resultant force vector in this case which is what I think might have confused you. It might be more clear if you consider the weight force as having 2 components mgsinθ and mgcosθ

The second question;
We'll call the skateboarder S, the walker W and the ground G.
Vsw = Vsg - Vwg
(the velocity of the skateboarder relative to the walker is equal to the vector sum of the velocity of the skateboarder relative to the ground - the velocity of the walker relative to the ground)
We are told what Vsw and Vsg are 4.24ms^-1 North West and 3ms^-1 North. From here use vector addition and see if you can get the answer.