Constants Part A You are lowering two boxes one on top of the other down the ram
ID: 1585239 • Letter: C
Question
Constants Part A You are lowering two boxes one on top of the other down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp Both boxes move together at a constant speed of 11.0 em/s The coefficient of kinetic friction between the ramp and the lower box is G 423, and the coefficient of static friction between the two boxes is 0 755 What force T do you need to exert to accomplish this? You may want to review (Pages 142-150) For help with math skills. you may want to review Besoling Vector Components Submit Renuest Answer For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Part B What is the magnitude f of the friction force on the upper box? Figure f= Subme Resuest Anseer Part C 250 hat is the dvection of the fricthion farce on the upper box? up the remp down the ramp SubenitExplanation / Answer
A) What force do you need to exert to accomplish this?
As there is no acceleration, the rope tension plus the ramp friction will balance the weight component parallel to the ramp.
T + (m + m)gcos = (m + m)gsin
T = (m + m)gsin - (m + m)gcos
T = (m + m)g(sin - cos)
tan = 2.50/4.75
= 27.76º
T = (48.0 + 32.0)9.81(sin27.76 - 0.423cos27.76)
T = 71.77 N
B) What is the magnitude of the friction force on the upper box?
The maximum friction force available is
Ff = mgcos
Ff = 0.755(32.0)9.81cos27.76
Ff = 209.73 N
However, as friction is a reactionary force, that is, it does not arise unless an active force opposes it, we need to check the weight component of the upper box acting parallel to the ramp
F = mgsin
F = 32.0(9.81)sin27.76
F = 146.2 N
as this is less than the maximum capable of being supported, no sliding between the boxes occur and the friction force balances the parallel weight component
F = 146.2 N
C) What is the direction of the friction force on the upper box? (up ramp or down)
Up the ramp (it keeps the upper box from sliding faster than the lower box)