Why is using Newton's First Law failing me?

[Murtuza Tipu]

A factory worker pushes a 29.7kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 32∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

What magnitude of force must the worker apply to move the crate at constant velocity?

So because constant velocity
∑ f=0
F_x *cos(360-32) -f_k = 0
F_x *cos(360-32) -(mg)(24) = 0
F_x *cos(328) - 69.8544N = 0
F_x = 69.8544N/cos(328∘)
F_x = 82.3708N

That turns out to be wrong, the expected answer is f_k the force of kinetic friction. Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer?

I also tried to find the square root of F_x² + F_y² but that didn't help either.

 

[paisiello2]

You didn't use the right friction force. Start with the definition of friction and maybe draw yourself a free body diagram.

 

[vela]

A factory worker pushes a 29.7kg crate a distance of 5.0m along a level floor at constant velocity by pushing downward at an angle of 32∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

What magnitude of force must the worker apply to move the crate at constant velocity?

So because constant velocity
∑ f=0
F_x *cos(360-32) -f_k = 0

I'm going to just use ##32^\circ## below; it doesn't make a difference. Your equation isn't correct. You should have
$$F_x - f_k = 0$$ or
$$F\cos 32^\circ - f_k = 0.$$ The x-component of F is ##F_x = F\cos 32^\circ##. Writing ##F_x \cos 32^\circ## doesn't make logical sense.

Why? Shouldn't I be able to use ∑F=0 in this problem to find the answer?

Yes, you can. Your thinking here is correct. It's just your execution is flawed for the reason paisello2 has hinted at.

 

[dauto]

You will save a lot of time and grief if you draw a free body diagram. have you done that?

 

[HalfLostAndSearching]

Using Newton's First Law, which states that an object will remain at rest or in constant motion unless acted upon by an external force, may not be applicable in this situation because there are multiple forces acting on the crate. The worker is applying a force at an angle, which means there is a component of that force in the horizontal direction, as well as the force of kinetic friction acting in the opposite direction. In order to use ∑F=0, all forces must be acting in the same direction.

In this scenario, the force of kinetic friction is the only force acting in the horizontal direction, so it is the only force that can be equated to the horizontal component of the worker's force. The vertical component of the worker's force is not relevant in determining the magnitude of force needed to move the crate at constant velocity.

Therefore, the correct approach would be to equate the horizontal component of the worker's force to the force of kinetic friction, as shown in the calculations above. This will give the correct magnitude of force needed to move the crate at constant velocity.