The last caller on today’s show asked for help studying for a physics test. The question was, a person gets their car stuck in the mud, ties the back of the car to a tree with a rope, and pushes horizontally on the rope with a force of 300 Newtons, if the angle of the rope is 5 degrees from straight, what is the force on the car?
Tom and Ray’s were right that it was a vector problem but their solution wasn’t right.
The whole point of attaching the rope to the tree and pushing was to increase the amount of force applied to the car over the 300 Newtons the person can apply. Tom and Ray answered 300N*cos(5) but cos(5) is less than one so this doesn’t give the stuck driver any mechanical advantage. The solution to the problem is to draw the diagram and remember SOHCAHTOA. in other words Sin (angle) =opposite over hypotenuse. the tension from the rope forms the force vector that actually pulls the car. So to arrive at the answer simply use Sin(5 degrees)=300N/(the tension on the rope). You should be able to solve it form there. Good Luck!
Yah. what he said. I wish I could draw a picture here but any how the 300 is side opp of the 5 Deg. sin=SO/HYP OR HYP=SO/SIN(5)=300/0.087155743=3442.11N
Your answer is correct but I’m not following your logic. Just remember the saying
sin= Oscar Had
cos=A Harry
tan=Old Aunt (Ok not Aunt but I felt like I should clean it up for here)
sin=Opposite/Hypotenuse
cos=Adjacent/Hypotenuse
tan=Opposite/Adjacent
The answer is the applied force divided by the sin of the angle as you did, not the applied force times the sin of the angle as stated on the show, and certainly not the applied force times the cos, also stated on the show. If they wanted to multiply instead of divide, the would have to multiply by the cosecant.
The above answers are not quite correct, because they neglect the fact that both the car and the tree are resisting the force applied by the student.
Here’s a picture (angles are exaggerated):
student
F
|_______
/ heta
/
car T T tree
The student pulls (or pushes) sideways on the rope with force F=300 N. In order to resist this force, the car and the tree both pull back on the rope with a force equal to the tension T. But since the resisting force is applied at an angle theta=5 degrees, only part of the force is in the sideways direction and able to counteract the force applied by the student, namely T sin(theta) for the car and T sin(theta) for the tree. This must be equal and opposite to the force F (the total force on the point must be zero, since everything is at rest), so we have
F=2T sin(theta).
An easy way to check that sin(theta) is correct is to imagine that theta = 90 degrees. Then both the tree and the car are pulling directly back with tension T, so in this case F=2T, which agrees with our formula above, since sin(90 deg) = 1.
Since the tension T is pulling on the car, the total force on the car is
T = F / (2 sin(theta)) = 1720 N.
I like your explaination, just one thing “The sdudent pulls (or pushes) sideways on the rope” I get confuse, One of our more recent (recently living) Generals once said (I believe it was Patton) “you can’t push a rope”.
That’s true, you can’t get anything useful done by pushing on the end of a rope 
But if you secure both ends, you can stand somewhere in the middle and push sideways on it. You get the same result as if you stood on the other side of the rope and pulled, so it really doesn’t make any difference. For example, clothes hanging on a clothesline pull it downward, and a bird sitting on the clothesline pushes it downward.
zzzzzzz/I
hype / Iside oppesite
-car/__I
-----^theta
not sure where that formula came from and it mite even be rite but theta is in wrong spot.
The theta can go in different places…here’s a bigger picture showing that. (Although, as usual in physics, accuracy is the enemy of clarity):
student
______________|F____________________
theta/| heta
/ |
/ |
/ |
T/ | T
/ |
/ |
______/theta__|__theta________
car middle tree
(We need to use the “code” tag to get the ascii art to work properly.)
All of the angles marked theta are equal. The picture shows that the component of the force exerted by the tree along the direction that the student is pulling (student-middle) is T sin(theta), since (student-middle) is opposite theta and (student-tree) is the hypotenuse. The force exerted by the car along the (student-middle) direction is also T sin(theta), since (student-car) is also the hypotenuse. So the total force resisting F is 2T sin(theta).
I took physics last semester, and already forgot half of it. However, the mytudor site might be right. I used cramster.com, where they solve the odd problems for almost all problems, and the whole book if you plunk out 40 bucks a year. Just a suggestion.
If you divide half of the perpendicular force (one half of the perpendicular force is being applied to the tree) by the sin of the angle you get the tension in the rope. You should divide the perpendicular force by the tan of the angle. However, every physicist knows that the difference between sin and tangent at small angles is negligible so the point is moot.
If you take the angle to extremes, i.e. 90 degrees, while applying the same 300 newton force you would arrive with 150 newtons pulling on the stuck car dividing by sin 90 = 1 yielding a longtitudinal pull on the car of 150 newtons. But dividing the lateral pull of 150 newtons the tan 90 = infinity you get 0 newtons of longitudinal pull. You are just trying to drag the car and tree laterally plus the rope will be infinitely long.
This takes me back to the days of statics.
Oops, yeah. I was wrong. The person below me had it right. I forgot the take the tension of the rope into account.
I’m pretty sure you don’t divide by 2(the force) for the 90 deg thing picture it as up with a pulley. if you lift a weight via pulley the same force is on the attachments both at the to0p and where it’s tide off.
another way to look at it is as a simple machine. you’d need to know lengths but then you could estimate the mechanical advantage by seeing the ratio of the change in hypotenuse length to a small deflection in the center point( the base of the triangle) it should approach a limit and that would be the secant probably.
No, I think we are right.
If you tell me exactly where you disagree, I can probably explain it better.
It might help to think of the theta=90 deg. case, and imagine that we’re trying to lift the objects using the rope, instead of pulling them. Then, we’re holding the middle of the rope, and a there are weights W attached to each end of the rope. The total weight is 2W, and we’ve got to support this weight with the force F holding up the middle of the rope. The total force up has to equal the total force down. So F = 2W.
Heres a way to break down the problem. Lisa, the physics student holds the middle of the rope. Weight lifters Abe and Ben each pull south with 150 newtons against Lisa. Lisa has to supply 300 newtons to the north or be pulled south. Abe and Ben now pay out rope while maintaining the 150 newton southward pull. When they reach the car’s bumper and the tree, Abe is pulling on the rope with 1721 newtons of force – 150 newtons south and 1715 newtons west and Ben is pulling with 1721 newtons on the rope – 150 newtons south and 1715 newtons east. Lisa is still pulling with 300 newtons north. The north-south forces balance and the west-east forces balance.
Abe attaches the rope to the the car bumper while Ben winds the rope around the tree. The car now begins to move. When Lisa’s mechanical advantage drops, she moves to the line between the car bumper and tree; has Ben take up the slack; and starts pulling north again.
Abe and Ben now realize that if they both can pull with 1721 newtons on each end of the rope, together they can apply 3442 newtons to the bumper of the car. So why all this fuss with the tree and the rope.
Eventually Abe and Ben get tired of this nonsense; grab the putative VW Bug by the bumpers; lift and walk it back onto solid ground. The ethical/moral questions is does Lisa invite Abe, Ben, both, or neither back to the dorm. Maybe working on this physics problem has given her a headache LOL.
I see your point now. I mis-read your formula. It looks like F1 in italics instead of F? so I really didn’t read it very closely.
The saying was “you can’t push a rope uphill.” You can certainly push it sideways.