Download 46) A furniture crate of mass 60

46) A furniture crate of mass 60.8 kg is at rest on a loading ramp that makes an
angle of 25.8 degrees with the horizontal. The coefficient of kinetic friction between
the ramp and the crate is .272. What force, (in Newtons) applied parallel to the
ramp, is required to push the crate up the incline at a constant speed?
Please go through the attached PDF document to understand about the horizontal and
perpendicular components on a body placed on an inclined plane. Also, refer a book like
the Physics volume 1 by Resnick and Halliday to understand how to draw free body
diagram of any such body (you can see this in the attached file also). In all problems we
assume that the value of g = 9.8 m/s2.
In this problem, we have the net acceleration on the body as zero since the body is
moving with a constant speed. So we have the equation for the forces parallel to the
inclined plane as
F – {f + mg Sin ()} = 0. It can be seen that the force F is what we
apply just to overcome the weight of the body mg Sin () and the force of static friction f
acting down the incline against F.
Again, there is no acceleration perpendicular to the incline, so we have the normal
force,
N = mg Cos ()
F = k N + mg Sin () = 0.272* mg Cos () + mg Sin ()
= 0.272* 60.8*9.8* Cos (25.8) + 60.8*9.8* Sin ()
= 405. 2414 N
-----------------------------------------------------------------------------------47) A shopper pushes a 40.7-kg cart up a 23.1-degree smooth ramp, what force (in
Newtons), directed along the incline, must the shopper exert to give the cart an
acceleration of .726 m/s/s?
If F is the force required to push the cart up with acceleration a, then the net force parallel
to incline is
F - mg Sin () = ma,
Since the force of friction is zero on a smooth ramp. So, we have
F = ma + mg Sin ()
= 40.7*0.726 + 40.7*9.8* Sin (23.1)
= 186. 0358 N
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48) A box slides down a 33.1-degree ramp with an acceleration of 1.3 m/s/s.
Determine the coefficient of kinetic friction between the box and the ramp. Enter
your answer accurate to the fourth decimal place.
The forces on the block are:
1. mg downward by the earth (gravity)
2. N the normal force by the incline and
3. f up the plane (force of friction) by the incline.
Taking the components parallel to the incline and writing Newton’s second law,
mg Sin () – f = ma, that is
f = 1.31 m + mg Sin (33.1).
There is no acceleration perpendicular to the incline, so we have
N = mg Cos (33.1).
Also we have f = k N.
Hence, k = (1.31 m + mg Sin (33.1)) / mg Cos (33.1)
= (1.31 + 9.8 Sin (33.1)) / 9.8 Cos (33.1)
= 0.81146 is the answer.
-----------------------------------------------------------------------------------------------------------49) A box weighing 5544 N is being pulled up an inclined plane that rises 2.22 m for
every 7.81 m of length measured along the incline. If the coefficient of friction is
.599, determine the force (in Newtons), applied parallel to the incline, necessary to
move it up the incline at constant speed.
The required force in this case will be F = mg Sin () + f
= mg Sin (k N
= mg Sin (k mg Cos (
The angle of the inclination can be found by, Sin-1 (2.22/7.81) = 16.51 degrees.
Hence F = 5544 Sin (16.51) + 0.599 * 5544 * Cos (16.51) = 4759.76 N
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50) A crate is held at rest on a frictionless 53.7-degree ramp by a rope parallel to the
incline. If the tension in the rope is 7061 N, then what is the mass (in kg) of the crate?
Taking the components of the force parallel to the incline, we have T - mg Sin
(when the crate is held at rest. Since the surface is frictionless, we have the tension
T = mg Sin (Therefore,
m= T/ g Sin (Sin (kg
-----------------------------------------------------------------------------------------------------51) A 40.6-kg wagon is towed up a hill which is inclined at 18.5 degree with respect to
the horizontal. The tow rope is parallel to the incline and has a tension of 145 N in it.
Assume that the wagon starts from rest at the bottom of the hill and neglect friction. How
fast (in m/s) is the wagon going after moving 67.2-meters up the hill?
If the initial velocity is zero as given, we have the equation for velocity:
v2 = v02 + 2ax = 2ax for one dimensional motion. Now using the
formulas we used above, we can see that the net force is
T - mg Sin (ma, which gives
a= T/m - g Sin ( Sin (m/s
Hence the velocity after the wagon travels a distance of 67.2m can be obtained by taking
the square root of the following equation which gives:
v2 = 2ax =2*0.4618*67.2 = 62.066 giving v = 7.8782 m/s
-----------------------------------------------------------------------------------------------------------54) A 31.9-kg block starts from rest on the top of a 33.6-degree incline and slides a
distance of 2.0 meters in 4.0 seconds. Determine the coefficient of friction between
the block and the incline. Enter your answer accurate to the fourth decimal place.
In this problem it can be seen that the forces acting parallel to the incline are the parallel
component of weight, the force of kinetic friction and the force due to the acceleration of
the body. So we have the net force as:
mg Sin (fk = ma.
Since the body starts from the top and is initially at rest, we have the equation for
displacement as x = (½) at2, giving the value of as
a= 2x/ t2 = 2*2/16 =0.25 m/s
Hence we have fk = mg Sin (ma = 31.9*9.8*Sin (33.6) – 31.9*0.25
= 165.0263 N
From the vertical component of the forces, we have
fk = k N = k mg Cos ().
Hence, k = fk / mg Cos () = 165.0263 / 31.9 * 9.8 *Cos ()
= 0.6338 is the answer.
One can use substitution rather than calculation at each stage by
keeping the numerical calculation only towards the end of the answer. Just
substitutes the equations and calculate the values only at the last stages. All
the best! Use no other fonts other than that used in this file since that may
change the Greek letters used.
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