Finding force with trig functions

[Pencil]

A shopper pushes a 7.5 kg shopping cart up a 13 degree incline. Find horizontal force needed to give the cart an acceleration of 1.41 m/s^2.

I know the answer is 28N, but I haven't yet figured it out.
I did F=ma F=(7.5)(1.41)cos13= 10.30
what didn't I do?
thanks

 

[BishopUser]

You seem to be forgetting that there are multiple forces acting on this cart and that F=ma means net force (sum of all forces) = ma

The way I do this problem is to create a coordinate system in which the x-axis is parallel to the slope of the plane, that way we can discount forces that are acting in the y-direction (we know that the sum of the forces in the y-direction must equal 0). Only 2 forces have x components, force of the person, and gravity.

Fnet = ma

F - (mgsin13) = ma
F = (7.5 kg)(9.8 m/s^2) + (9.8 m/s^2)(7.5kg)(sin13)
F = 27.109N

this force we calculated is the force needed to create that acceleration if the person was pushing parallel to the x-axis, but we know he wants to push this horizontally so he will require a little more force.

Fcos13 = 27.109
F = 27.8N

 

[kyle8921]

Your approach is correct, but you forgot to include the gravitational force acting on the shopping cart. The total force needed to accelerate the cart up the incline is the sum of the horizontal force and the force of gravity. So the correct equation would be F = ma + mgcos13, where m is the mass of the cart, a is the acceleration, g is the acceleration due to gravity (9.8 m/s^2), and 13 is the angle of the incline. Plugging in the values, we get F = (7.5)(1.41) + (7.5)(9.8)cos13 = 28N, as you correctly stated. Keep up the good work!