# Normal force

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#### Tokspor

##### Full Member

An older gentleman drags a wooden crate (10 kg) out of storage. The 120 N force that he applies on the crate makes an angle of 30 degrees with the horizontal. The coefficient of kinetic friction between the floor and the crate is 0.1. What is the frictional force on the crate?

A. 2.4 N
B. 3.8 N
C. 7.2 N
D. 9.8 N

The answer is B. According to the text, the frictional force is calculated by finding the normal force, which is supposed to be F(normal) = mg - F*sin(30).

I don't understand how this equation was set up. If the upward force is F*sin(30), shouldn't this only add to the F(normal)? So I would think that F*sin(30) = F(normal) - mg.

To me, having an upward force on the crate is sort of like a person in an elevator accelerating upward, where ma = F(normal) - mg.

#### loveoforganic

##### -Account Deactivated-
10+ Year Member
Technically, I guess you could see it that way. For the purposes of the MCAT though, I think the normal force is normally defined as the reactionary force provided by the surface the box rests on, not the force opposing gravity.

#### traitorman

##### Full Member
7+ Year Member
im trying to respond to your post with a picture i drew in paint but i cant seem to upload it because its too many kilobytes!

basically, if u divide the box up into the X and Y axis, The forces on the Y axis are Normal force + Fsin120 = mg.

Rearranging the forces: Normal force = mg - Fsin120

Friction force = Normal force x 0.1 = 3.8N

7+ Year Member

#### Tokspor

##### Full Member
im trying to respond to your post with a picture i drew in paint but i cant seem to upload it because its too many kilobytes!

basically, if u divide the box up into the X and Y axis, The forces on the Y axis are Normal force + Fsin120 = mg.

Rearranging the forces: Normal force = mg - Fsin120

Friction force = Normal force x 0.1 = 3.8N

Why is this not the case for upward acceleration in an elevator? The forces along the Y axis would be F(normal) + ma(elevator) = mg. So m(elevator) = F(normal) - mg.

However, the correct equation is ma(elevator) = F(normal) - mg.

---

The only difference I see is that there is no net upward acceleration in the case with the crate. So for cases with no net acceleration, any additional force in the same direction as the normal force helps to decrease what the normal force itself needs to match the downward force?

But when there is acceleration, any force in the direction of the normal force only adds to it?

#### jpatel

##### Membership Revoked
Removed
7+ Year Member
you are pushing down with the books with the force of mg

The dude is pushing up the books by Fcos.

So "technically" if ur the ground, you would only feel the force of
mg - Fcos.

You can think of normal force (atleast it helps me) as how hard ground is feeling the force lol.

#### IntelInside

##### Full Member
10+ Year Member
7+ Year Member
Why is this not the case for upward acceleration in an elevator? The forces along the Y axis would be F(normal) + ma(elevator) = mg. So m(elevator) = F(normal) - mg.

However, the correct equation is ma(elevator) = F(normal) - mg.

---

The only difference I see is that there is no net upward acceleration in the case with the crate. So for cases with no net acceleration, any additional force in the same direction as the normal force helps to decrease what the normal force itself needs to match the downward force?

But when there is acceleration, any force in the direction of the normal force only adds to it?

You are forgetting that you have to specify + and - directions because force is a vector. If i specify up as + and down as neg for the elevator example you gave, we can then see that:

ma(elevator)=Fn-mg

whereas what you did was:

ma(elevator)=mg-Fn we are contradicting + and - here Fn would have to be positive too because it points in the upwards direction. However, this formula would work if it was downward acceleartion and I specify down as + and up as -

#### UCB05

##### Full Member
Why is this not the case for upward acceleration in an elevator? The forces along the Y axis would be F(normal) + ma(elevator) = mg. So m(elevator) = F(normal) - mg.

However, the correct equation is ma(elevator) = F(normal) - mg.

---

The only difference I see is that there is no net upward acceleration in the case with the crate. So for cases with no net acceleration, any additional force in the same direction as the normal force helps to decrease what the normal force itself needs to match the downward force?

But when there is acceleration, any force in the direction of the normal force only adds to it?

Has the question been answered thoroughly for the OP? If not, here's my two cents~

Normal force is the force perpendicular to a surface required for an object to remain static in relation to that surface, correct? Thinking of it that way, you should sum all your forces up, and use that sum to find your normal force.

In the opening example, the vertical forces are gravitational downward: mg, and old man upward: Fsin30 (fyi 30-60-90 right triangle means upward force is 60). Therefore Fn = mg - Fsin30.

In an elevator, the upward force used to accelerate the system is applied to the ELEVATOR, not the object. Therefore, your relative acceleartion to the elevator which itself is accelerating upward is (g+a), and the normal force required to maintain your relative position in the accelerating elevator is actually Fn = m(g+a) = mg + ma. That is where the difference lies. In the question the force is being applied to the object itself. In the elevator the force is being applied to the elevator, not the object, so you require extra normal force to accelerate upward as well.