A 2000 pound car is parked on a street that makes an angle of 12* with the horizontal (think it's a street on a hill). a) find the magnitude of the force required to keep the car from rolling down the hill. b) Find the force perpendicular to the street.
a. Force to prevent car from rolling = v
v=<cos12°,sin12°>
F=projvF=(F•v/|v|^2)v= (F•v)v
<0,2000>•<cos12,sin12>= 0+2000sin12 = 415.82lbs
b. Find the angle of the perpendicular in relation to the force of 2000lbs. To do this, I drew a triangle, one line horizontal, one line being the 12° angle between it and the horizontal, and one being perpendicular to the 12° line(90°). Connect the lines and the third angle should be 78°.
Then you do the same as the last.
Force to prevent car from rolling(perpendicular) = c
c=<cos78°,sin78°>
F=projcF=(F•v/|v|^2)v= (F•v)v
<0,2000>•<cos78,sin78>= 0+2000sin78 = 1956.3lbs
mass = 2000lbs * 0.454kg/lb = 908kg.
Fc = mg = 908kg * 9.8N/kg = 8898.4N @
12deg.
a. Fp = 8898.4*sin(12) = 1850.4N. = Force parallel to the street = Force required to keep the car from roling down hill.
b. Fv = 8898.4 * cos12 = 8704N = Force
perpendicular to the street.
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a) Well, the car certainly weighs a lot, but it's not very athletic when it comes to rolling down hills. To find the force required to keep it from rolling, we need to consider the component of its weight that acts downhill. This can be calculated using some trigonometry.
Let's start by finding the component of the weight that is parallel to the street, which will be the force trying to make it roll. We'll call this force F_roll. The total weight of the car is 2000 pounds, but we'll need to convert that to the appropriate unit of force, which is pounds force (lbF). Since weight is mass times gravity, and the acceleration due to gravity is approximately 32.2 ft/s^2, we can calculate the weight of the car as follows:
Weight = mass * gravity = (2000 lb) * (32.2 ft/s^2) = 64,400 lbF
Now, let's find the component of this weight that acts downhill. We can use the sine of the angle of the street (12°) to determine this:
F_roll = Weight * sin(angle)
F_roll = 64,400 lbF * sin(12°)
F_roll ≈ 13,220 lbF
So, the magnitude of the force required to keep the car from rolling down the hill is approximately 13,220 pounds force.
b) Next, let's find the force perpendicular to the street, which will be the component of the weight acting in that direction. We'll call this force F_perpendicular. Again, we can use trigonometry to determine this.
F_perpendicular = Weight * cos(angle)
F_perpendicular = 64,400 lbF * cos(12°)
F_perpendicular ≈ 63,130 lbF
So, the force perpendicular to the street is approximately 63,130 pounds force. Just remember, the car won't be clowning around when it comes to staying straight on that hill!
To find the magnitude of the force required to keep the car from rolling down the hill, we need to consider the weight of the car and the force of friction acting on it.
a) The weight of the car can be calculated using the formula:
Weight = mass × gravity
Given that the weight of the car is 2000 pounds, we need to convert it to mass. We know that 1 pound is equivalent to 0.454 kg (approximately). Therefore, the mass of the car is:
Mass = 2000 pounds × 0.454 kg/pound
Next, we need to calculate the force of gravity acting on the car:
Force of Gravity = Mass × Gravity
The acceleration due to gravity is approximately 9.8 m/s².
Now we can find the component of the weight of the car that acts down the hill. This component can be calculated using the formula:
Force down the hill = Weight × sin(angle)
Here, the angle is given as 12 degrees, but we need to convert it to radians:
Angle (in radians) = Angle (in degrees) × π/180
Finally, the magnitude of the force required to keep the car from rolling down the hill is equal in magnitude but opposite in direction to the force down the hill. So the answer is the negative of the force down the hill.
b) The force perpendicular to the street can be calculated using the formula:
Force perpendicular = Weight × cos(angle)
Using the same angle as before (12 degrees), but this time not converted to radians, we can substitute the values and calculate the force perpendicular.