Question

How much force is required to drag a 90 lb. box up this "frictionless" inclined plane?

109 lb.
10 lb.
81 lb.
9 lb.

Answer 1

i would have to say a

Answer 2

hmmmmmm can you explain how can i find this answer?

Answer 3

Hi, Manny!
Only three forces are important here: 1) the weight of the box; 2) the pull of gravity in the direction downward and parallel to the sloping surface, and the unknown force.

We aren't given the angle of this ramp; but are given the lengths of the sides of the triangle. Could you figure out the angle from the side lengths and one of the inverse trig functions?

Answer 4

i don't understand tbh.. i need to figure out how much force will it take to drag a 90lbs box up the incline plan... I'm taking a physical science test

Answer 5

Have you drawn a free body diagram before?

Answer 6

no

Answer 7

Have you seen any sample problems where you have to split a single force into two component forces?

Answer 8

I'm trying to figure out where you're coming from in terms of understanding the mechanics of this problem, so that I can build upon what you already know.

Answer 9

idk anything i just want to finish this test that way i can take my sorry retricehome and drink my sorrows away

Answer 10

sorry a s s * home

Answer 11

I'll briefly outline the procedure: 1) find the measure of the smallest angle of this triangle; 2) the straight downward force on the box is 90 lb. Box can't fall straight down becasue it's on a ramp. The downward force can be split into two perpendicular forces, one of which is not straight down, but rather downward and parallel to the ramp.
Does this ring a bell at all?

Answer 12

this is an interesting topic, but if you're taking a test right now and if you haven't worked with free body diagrams and component vectors, it'd take some time to explain the problem fully.

Answer 13

todays my brothers birthday. he just turned 19, he passed away.... all day my minds been dwelling off topic...

Answer 14

My deepest sympathy. Don't see how you could manage to concentrate on math/physics at all on a day like this. So sorry!!

What do you want to do? Continue or come back here later, when you're feeling a bit better?

Answer 15

We can determine the angle that the slope makes with the horizontal using the inverse sine function. That angle would be \[\sin ^{-1}\frac{ H }{ 9H }=\sin ^{-1}\frac{ 1 }{ 9 }\]

Answer 16

This comes out to about 6 degrees.

Answer 17

there's always a reason to get back on your feet.. thank you lets continue. hmmmm

Answer 18

is it 10?

Answer 19

All right. Sorry again.

Does my use of the inverse sine function ring a bell with you? In the triangle given, the side opposite the angle 6 degrees is marked H, and the hypotenuse (length of the ramp) is marked 9H.

Answer 20

As a matter of fact, the component of that 90 lb in the direction opposite to the pull of the rope is 10 lb, and that asnwers this question. A 10-lb force will pull the box up this ramp if there's no friction.

Answer 21

So it wasn't even necessary to find the angle in degrees. What you did has worked.

Answer 22

thanks you you helped me!.

Answer 23

If I had to write out the answer formally, this is what I'd write:\[F _{e}=(90 lb)\sin \sin ^{-1}\frac{ 1 }{ 9 }=(90 lb) (\frac{ 1 }{ 9 }) = 10 lb\]

Answer 24

My sympathy, JMC. Hope you can remember the good times you enjoyed with your brother and cherish what you learned from him. Best to you. Over and out.

Answer 25

Bye, JMC.

Answer 26

thank you so much my brother Raymond was a MARSOC warrior for the USMC. Soon I'll avenge his death..... and pay a tribute for fallen heaven hound.. Rahhh! thank you mathmale

Answer 27

What an honor for Raymond to be in the USMC, and for you that he was your bro. Going into the USMC yourself?