A box of mass 40kg rests on a flat horizontal surface. It is pulled by a constant horizontal force of 200N and the frictional force acting between the box and the horizontal surface is 120N.
2006-08-22 06:45:36 · 10 answers · asked by jeanpace89 1 in Education & Reference ➔ Homework Help
You have not explained the problem clearly enough.
I believe you have been asked to determine the Coefficient of Friction.
Downward Force = mg (Calculate this)
Net Horizontal Force = (200 - 120)N
= 80N
Sketch the force triangle and calculate the value and direction of the Resultant Force and also, the Tangent of the Vertical and Net forces to determine the Friction Angle.
Tangent = Coefficient of Friction
Hint: The Resultant Force and Direction is that required to just overcome the frictional resistance.
2006-08-30 06:16:16 · answer #1 · answered by CurlyQ 4 · 0⤊ 0⤋
Stop cheating and do your homework. Whats the point of it if your just gonna slap it out on Yahoo Answers??
Think about it. One of the most key skills in mathematics, physics and life! .. is being able to resolve forces.
1) Think about motion. First up or down? Is the box moving up or down or not at all in these directions?? **Hint Newtons 2nd (F=ma .. in short) and 3rd laws of motion (To every action - or force in this case - there is an equal and opposite reaction - or again force in this case).**
2) Now think horizontally. Is the box (assuming a 2D problem), moving left or right? Well.. your given the forces. frictional force will act in the opposite direction to the applied force by definition at the interface between the box and the surface (ie where the two touch). One is bigger than the other. Now resolve this with a linear expression. (ie no squared ro greater power terms.. simply ie. a+b=c **Hint: Take the applied force as a positive figure and the frictional force, since its in exactly the opposite direction here, as a negative one.**
Even when facing angled forces which dont follow the longitudinal and lattitudinal directions (x & y axis or up-down & left-right), we can resolve these forces using simple geometry and in essence, The SOH CAH TOA method. If you don't know this don't worry. I'm sure your teacher will teach you it soon enough and it makes life VERY easy... even if it sounds a little scary and complex to start with.
Trust me.. I AM a rocket scientist :)
2006-08-22 14:12:56 · answer #2 · answered by frohike47 2 · 0⤊ 0⤋
The magnitude is the force applied less the resistance, in this case the horizontal pulling force less the horizontal surface frictional resistance. You give no direction so none can be determined other than to say in the direction of the horizontal pulling force.
2006-08-27 08:38:20 · answer #3 · answered by Magic One 6 · 0⤊ 0⤋
The magnitude would be only moderate, and the direction would be forward. The main thing is, don't you think you're a little young to be fretting your head about problems like this?
2006-08-22 13:55:25 · answer #4 · answered by yahoohoo 6 · 0⤊ 0⤋
E=mcsquared
2006-08-22 13:55:37 · answer #5 · answered by Marcus R. 6 · 0⤊ 0⤋
Fx= mg= 40*9.81= 392.4 kN
Fy= 200 - 120 = 80 N = 0.8 kN
finish the rest
2006-08-22 15:08:33 · answer #6 · answered by A.A 1 · 0⤊ 1⤋
i won't solve it, but i'll just give you a hint-
take the component of forces and find out, or better still, use vector algebra
2006-08-22 13:57:24 · answer #7 · answered by s_kundu88 3 · 0⤊ 1⤋
You really need to start doing your own homework!
2006-08-22 13:51:42 · answer #8 · answered by woohoo 3 · 1⤊ 1⤋
christ i left school so i wouldnt have to do stuff like this anymore
2006-08-22 13:50:59 · answer #9 · answered by KEV D 3 · 1⤊ 1⤋
thanks for letting me know how dumb i really am.
2006-08-29 23:02:53 · answer #10 · answered by Anonymous · 0⤊ 1⤋