A block weighing 87.5 N rests on a plane inclined at 25.0° to the horizontal. A
ID: 1491703 • Letter: A
Question
A block weighing 87.5 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.367 and 0.156.
(a) What is the minimum value of F that will prevent the block from slipping down the plane?
(b) What is the minimum value of F that will start the block moving up the plane? .
(c) What value of F will move the block up the plane with constant velocity?
Explanation / Answer
here,
weight , w = 87.5 N
m = w/g = 8.93 kg
theta = 25 degree
phi = 35 degree
us = 0.367
uk = 0.156
(a)
static frictional force , Ff = us * ( m * g * cos(theta) - F * sin(phi))
for the block to not to move
ff + F * cos(phi) - m * g * sin(theta) = 0
us * ( m * g * cos(theta) - F * sin(phi)) + F * cos(phi) - m * g * sin(theta) = 0
0.367 * ( 87.5 * cos(25) - F * sin(35) )+ ( F * cos(35) - 87.5 * sin(25)) = 0
F = 12.94 N
the minimum value of F that will prevent the block from slipping down the plane is 12.94 N
B)
for the block to move
static frictional force , Ff = us * ( m * g * cos(theta) - F * sin(phi))
for the block to not to move
- ff + F * cos(phi) - m * g * sin(theta) = 0
- us * ( m * g * cos(theta) - F * sin(phi)) + F * cos(phi) - m * g * sin(theta) = 0
- 0.367 * ( 87.5 * cos(25) - F * sin(35) )+ ( F * cos(35) - 87.5 * sin(25)) = 0
F = 64.18 N
the minimum value of F that will prevent the start the block moving up the plane is 64.84 N
c)
kinetic frictional force , Ff = uk * ( m * g * cos(theta) - F * sin(phi))
for the block to not to move
- ff + F * cos(phi) - m * g * sin(theta) = 0
- uk * ( m * g * cos(theta) - F * sin(phi)) + F * cos(phi) - m * g * sin(theta) = 0
- 0.156 * ( 87.5 * cos(25) - F * sin(35) )+ ( F * cos(35) - 87.5 * sin(25)) = 0
F = 54.31 N
the value of F will move the block up the plane with constant velocity is 54.31 N
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