Acart is pulled horizontally with a 156.6 n force to the right at a 67 degree angle with respect to the ground. the cart is moving at constant velocity across a rough surface. a. draw a free body diagram.

b. calculate the vertical and horizontal components of the applied force.
c. determine the horizontal friction acting on the cart (hint: the cart is moving at constant velocity.)
d. determine the normal force applied on the cart. remember that there is a vertical component of the applied force and that the cart is not accelerating

b) Vertical= 144.15N [up]

horizontal= 61.19N [right]

c)  -61.19N [Left]

D) 9.8m - 144.15 = Fn

Explanation:

a) to draw the free body diagram, you would draw a square to represent the object, being the cart. The forces acting on it are Force of gravity (Fg going down), Normal force (Fn going up), Force applied (Fa going up right), and Force of friction  (Ff,going to the left, opposite of Fa).

b) So the applied force is 156.6N, and has an angle of 67. the 156.6N is the hypotenuse, so you would need to find the opposite side (vertical component), and adjacent side (horizontal component):

vertical:

sinθ = opp / hyp

sin67 = opp / 156.6

opp = 144.15N [up]

horizontal:

cosθ = adj / hyp

cos67 = adj / 156.6

adj = 61.19N [right]

c)Since the cart has constant velocity, the acceleration is 0, meaning no net force (fnet = ma, a = 0, fnet = 0), so the horizontal component of the applied force is equal to the frictional force. Heres why mathematically. (X DIRECTION ) :

Fnet = Fa + Ff

0 = Fa + Ff

-Ff = Fa.

So the force of friction has the same magnitude, but opposite direction. So since the horizontal component of applied force was 61.19 N, the frictional force will be -61.19N

d) So now we are looking at the Fnet in the y direction, which is also 0 because the car is remaining on the ground, its not going up nor down:

Fnety = 0

Fnet is composed of Fg, Fn and Fay (force applied  in y direction);

fnet = Fn + Fay + Fg.      Fn and Fay are both up, and Fg is down, so u need to subtract Fg:

fnet = Fn + Fay - Fg.

0 = Fn + 144.15 - mg

mg - 144.15  = Fn

9.8m - 144.15 = Fn

Since it didnt give you a mass thats how much you could simplify it down to, in order to get the Fn

iron would move it is magnetic there is a mag field around the current in the wire

do you have a formula possibly