You are lowering two boxes, one on top of the other, down the ramp shown in the
ID: 2182940 • Letter: Y
Question
You are lowering two boxes, one on top of the other, down the ramp shown in the figure by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.460, and the coefficient of static friction between the two boxes is 0.806.
a) What force do you need to exert to accomplish this?
b) What is the magnitude of the friction force on the upper box?
c) What is the direction of the friction force on the upper box?
- up the ramp
- down the ramp
Explanation / Answer
angle of ramp, = arctan(2.5/4.75) =27.758 degrees
Now as blocks are moving with no acceleration they can be considered as a system
and N be the normal force which aplied by wedge on blocks
then perpendicular to slope of wedge,
a) N = (32+48)*9.8*cos
and along direction of slope
T+ *N = mg*sin
force applied on the block by person = 80*9.8*(sin27.758-0.46*cos27.758) [we are using coefficient of kinetic friction here) = 46.0002 N
b) Now upper box does not have any acceleration and only forces acting on it are,
gravity and friction by lower block slong the direction of slope
thus, Fr = mgsin = 32*9.8*sin27.758 =146.055 N
c) and direction on upper box will be up the ramp.
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