A(n) 1100 kg car is parked on a 4 degree incline. The acceleration of gravity is 9.8 m/s^2. Find the force of friction keeping the car from sliding down the incline. Answer in unites of N.
2006-11-27 11:27:44 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Draw a free-body diagram of the car. In the free-body diagram there are 3 forces acting on the car:
1) The force of gravity (Fg) acts straight downward
2) The normal force (Fn) acts perpendicular to the slope and makes an angle of 4 degrees with the vertical axis
3) The friction force keeping the car in place (Ff) acts parallel to the slope and up the slope, making an angle of 4 degrees with the horizontal axis.
Since the car is stationary the sum of the forces on the car is zero.
Make the positive x axis point up the slope and the positive y axis point up the normal force. This way the only force with an angle is the gravitational force (Fg). Now make an equation for the sum of the forces in the x direction:
sum(x) = 0 = Ff - Fg*sin(4)
You multiply Fg by sin(4) to get the x component.
Fg is equal to the force of gravity on the car, which is the mass of the car times the acceleration of gravity:
Fg = 1100 * 9.8 = 10780 N
Now solve the sum(x) equation for Ff to find the force of friction needed to keep the car from sliding down the incline.
Ff = Fg*sin(4) = 10780*sin(4) = 752 N
I hope this helps!
2006-11-27 11:29:41 · answer #1 · answered by Anonymous · 1⤊ 0⤋
The reactive forces must be equal. The weight of the car can be easily calculated, and it acts vertically downward. The force on the slope then reacts normal to the surface ( 4 degree from vertical), or cos 4 deg. The difference is a force parallel to the surface of the slope, or sin 4 deg. The force along that direction is the friction force. I'll let you do the math.
2006-11-27 11:34:18 · answer #2 · answered by maddojo 6 · 0⤊ 1⤋
The horizontal force is proportional to the sine of the angle, and the vertical force to the cosine. The horizontal force is 1100 kg x 9.8 m/s^2 x sin(4 degrees); the sine is 0.0697.
2006-11-27 11:32:28 · answer #3 · answered by Anonymous · 0⤊ 1⤋
The force pulling on the car is
Sin(4)*m*g
Since the car is staionary, this is offset by friction
j
2006-11-27 11:35:54 · answer #4 · answered by odu83 7 · 0⤊ 1⤋
Fw=mg
Fw=(1100kg)(9.8m/s^2)
Fw=10780N
Fw(x)=751.9N
since Fw(x) is 751.9N, Ff is also 751.9N
2006-11-27 11:42:57 · answer #5 · answered by 7 · 0⤊ 0⤋