Can Physics Solve These Real-World Problems?

[Nanabit]

1. A student gets his car stuck in a snow drift. Not at a loss, having studied physics, he attaches one end of a stout rope to the vehicle and the other end to the trunk of a nearby tree, allowing for a small amount of slack. The student then exerts a force F on the center of the rope in the direction perpendicular to the car-tree line. If the rope is inextensible and if the magnitude of the applied force is 491 N, what is the force on the car? (Assume equilibrium conditions.)

I'm very very bad at summing up the forces. I know that the sum of all forces in both the x and y directions is zero, and the same with torque, but honestly I don't know where to start because I have no experience in this type of problem. Same with #2.

2. A vertical post with a square cross section is 11.0 m tall. Its bottom end is encased in a base 1.50 m tall, which is precisely square but slightly loose. A force 5.60 N to the right acts on the top of the post. The base maintains the post in equilibrium.
Find the force which the top of the right sidewall of the base exerts on the post. (to the left and to the right.) b. Find the force which the bottom of the left sidewall of the base exerts on the post. (to the left and to the right.)

3. A carpenter's square has the shape of an L (d1 = 16.0 cm, d2 = 4.00 cm, d3 = 4.00 cm, d4 = 11.0 cm). Locate its center of gravity. (Hint: Take (x,y) = (0,0) at the intersection of d1 and d4) (d1 is the length of the long part of the L, d2 is the width of it, d4 is the length of the bottom part of the L, and d3 is the width of it.)

I know what the formula is for this, but I am a bit confused because I don't have any masses. Plus, are the x value and y values in the formulas just the distances in the x and y directions??

4. For safety in climbing, a mountaineer uses a 45.0 m nylon rope that is 10.0 mm in diameter. When supporting the 75.0 kg climber on one end, the rope elongates by 1.80 m. Find Young's modulus for the rope material.

y = FLo/change in L A F= 75 kg(9.8 m/s^2); L = 45.0 m; change in L = 1.8 m; A = pi r^2 = 7.85x10^-5. What am I missing?

Thanks. Sorry I'm so clueless.

 

[mani]

he attaches one end of a stout rope to the vehicle and the other end to the trunk of a nearby tree, allowing for a small amount of slack. The student then exerts a force F on the center of the rope in the direction perpendicular to the car-tree line. If the rope is inextensible and if the magnitude of the applied force is 491 N, what is the force on the car? (Assume equilibrium conditions.)

The vertical "sag" and the angles formed, when the rope just becomes taut, will depend on the value of the "a small amount of slack".

The tension of the rope when taut will in turn depend on the angle between the rope and the horizontal

 

[HallsofIvy]

1. A student gets his car stuck in a snow drift. Not at a loss, having studied physics, he attaches one end of a stout rope to the vehicle and the other end to the trunk of a nearby tree, allowing for a small amount of slack. The student then exerts a force F on the center of the rope in the direction perpendicular to the car-tree line. If the rope is inextensible and if the magnitude of the applied force is 491 N, what is the force on the car? (Assume equilibrium conditions.)

You can't get a final numerical answer without know both the distance from the car and the "amount of slack" or, at least, the ratio between them.
Draw a picture showing the rope from the car to the tree, then after being pushed a small amount perpendicular to the line. You have two right triangles. You know the force perpendicular to the line is 491 N and you know the "force triangle" is similar to the triangle formed by the rope. The force on the car, divided by 491 N is equal to 1/2 the distance from the car to the tree divided by the "amount of slack".

For example, if distance between the car and the tree is 100 feet and he leaves 0.1 ft slack (1.2 inches), the ratio is 50/0.1= 500. The force on the car is 500*491 N.

Notice that the less slack, the greater the multiplier, yet there must be some non-zero slack for this to work!