Normal force and acceleration of 7 kg block on 20° frictionless incline

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A horizontal force of 73 N pushes a 7 kg block up a frictionless incline that makes an angle of 20° with the horizontal.
(a) What is the normal force that the incline exerts on the block?
64.5N
from cos(20)*68.67

(b) What is the acceleration of the block? (Hint: assume up the block as the positive direction.)
3.35m/s2

from sin(20)*68.67 = 23.48
F = MA
23.48/7 = 3.35
 
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What about that horizontal 73 N force? Seems like you forgot all about it.
 
The 73N would affect the acceleration of the block right? I don't see how it affects the normal force.

So should it be 73N - 23.48N = 48.52N.
And plug that in the F = MA?
 
maniacp08 said:
The 73N would affect the acceleration of the block right? I don't see how it affects the normal force.

So should it be 73N - 23.48N = 48.52N.
And plug that in the F = MA?

That 73N force acts at an angle to the block. not parallel to the incline so there will be force transferred normally to the slope.
 
maniacp08 said:
The 73N would affect the acceleration of the block right? I don't see how it affects the normal force.
Since the 73N force is horizontal (and the incline is not) it will affect both the acceleration and the normal force.

Find the components of the horizontal force parallel and perpendicular to the incline.