A 50-kg cart is on an incline of 30 degrees above the horizontal. The cart is at rest. The static coefficient of the slope is 0.85 and the kinetic coefficient is 0.50. What is the force due to friction on the cart?

The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude w), the normal force (mag. n), and the friction force (maximum mag. f ).

In the horizontal direction, we have

n cos(120º) + f cos(30º) = 0

-1/2 n + √3/2 f = 0

n = √3 f

and in the vertical,

n sin(120º) + f sin(30º) + (-w) = 0

n sin(120º) + f sin(30º) = (50 kg) (9.80 m/s²)

√3/2 n + 1/2 f = 490 N

Substitute n = √3 f and solve for f :

√3/2 (√3 f ) + 1/2 f = 490 N

2 f = 490 N

f = 245 N

(pointed up the incline)

co2reflects heat radiation back to earth buning coal and oil emits co2 globe warms