Physics-Friction:push or held back?

[Physics_girl1]

Homework Statement

Question:
A safe weighing 2000 N is to be lowered at constant speed down skids 4m long, from a truck 2m high

a) If the coefficient of sliding friction between safe and skids is 0.30, will the safe need to be pulled down or held back?

b) How great a parallel force to the skids is needed?


i really need help with this my teacher didn't go over it with us

 

[Kurdt]

What have you tried so far with the problem or what do you know about kinetic (or sliding) friction.

 

[Physics_girl1]

i know that f=\mu*normal force and that the normal force is equal to mg, which is 2000N so kinetic force is 600N but i really don't understand the problem all that well

 

[Kurdt]

Well basically the safe is on an incline (the skids) which forms part of a right angled triangle with hypotenuse the legth of the skids and one of the sides the height of the van. So this problem also considers components of force.

What you need to do is find out whether the box can slide down the ramp by itself because the component of its weight down the slope is greater than the friction it encounters on the slope. Remember that on the slope the friction will be given by the component of the weight in a particular direction perpendicular to the slope.

 

[Physics_girl1]

ok thanks i will try that!

 

[Physics_girl1]

ok i spoke to soon. so what i should do is add the components of force together? I'm sorry I'm askin so many questions we just started Newton's laws and friction/inclines this week and i think the test is next week

 

[Kurdt]

First you need to resolve the weight of the safe down that particular incline. That means you need to work out the angle of inclination. I'd suggest drawing a diagram of the situation of you don't have one to help you picture what is going on.

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#vec5

The above link has some help on resolving vectors.

 

[Physics_girl1]

thanks i'll check out the site and then later post my answer

 

[Physics_girl1]

ok so i got theta as 23.59, force as 1833.03, normal force as 800N, and sliding friction as 240. does it look correct?

 

[Kurdt]

Your angle doesn't look correct to me. Try again.

 

[Physics_girl1]

i keep getting the same thing. when i take the inverse sine of 2/5 i get 23.58 and when i take the inverse cosine of (root 23)/5 i still get 23.58 and that's the angle of elevation from the skids and the ground

 

[Kurdt]

Ok. Are the skids 4m long or 5m?

 

[Physics_girl1]

okay i see my mistake now, sorry the skids are 4m my new answers are theta as 30, force as 1732.05, normal force as 1000N and kinetic friction as 300 is it correct now?

 

[Kurdt]

I've got the force down the slope as 1000N and the normal as 1732N. Once you find the friction force you can answer the first question.

 

[Physics_girl1]

could you explain please? i have the right numbers but for the wrong forces

 

[Kurdt]

The component down the slope forms the opposite side of a right angle triangle of forces. The weight being the hypotenuse and the normal force being the adjacent side. Therefore the force down the slope is the weight multiplied by the sine of the angle. The normal is the weight multiplied by the cosine of the angle.

 

[Physics_girl1]

oh, ok thanks for your help.and letter b is just asking how many Newtons are required to have to push the safe, right?

 

[Kurdt]

For part two you just need to find out the extra force that needs to be applied to equal the friction so the safe moves at a constant speed.

 

[Physics_girl1]

so you would subtract the kinetic fricton from the force of the slope?

 

[Kurdt]

Yes kinetic friction subtracted from force parallel to the slope.

 

[Physics_girl1]

ok thanks a bunch i understand it so much better now