Resolving Forces in an inclined plane ( Mechanics )
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I have exams after a few days and I'm doing all I can to understand the concept of resolving forces. With hard luck and a few hours of devotion, I acquired basic knowledge on Resolving Forces and was able to solve almost all questions and then this one came out.
A car of mass 850 kg is travelling, with acceleration 0.3m(s^-2) up a straight road inclined 12 degrees to the horizontal. There is a force resisting the motion of 250 N. Calculate the magnitude of the driving force.
Please help me out on this one. I'm really confused.
Plus, if you have great resources that can help a layman understand Resolving Forces to its depth, please add them too.
Thanks in Advance
classical-mechanics
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
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add a comment |
$begingroup$
I have exams after a few days and I'm doing all I can to understand the concept of resolving forces. With hard luck and a few hours of devotion, I acquired basic knowledge on Resolving Forces and was able to solve almost all questions and then this one came out.
A car of mass 850 kg is travelling, with acceleration 0.3m(s^-2) up a straight road inclined 12 degrees to the horizontal. There is a force resisting the motion of 250 N. Calculate the magnitude of the driving force.
Please help me out on this one. I'm really confused.
Plus, if you have great resources that can help a layman understand Resolving Forces to its depth, please add them too.
Thanks in Advance
classical-mechanics
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
$endgroup$
$begingroup$
Is the force resisting the motion in addition to gravity?
$endgroup$
– DJohnM
May 6 '16 at 18:20
add a comment |
$begingroup$
I have exams after a few days and I'm doing all I can to understand the concept of resolving forces. With hard luck and a few hours of devotion, I acquired basic knowledge on Resolving Forces and was able to solve almost all questions and then this one came out.
A car of mass 850 kg is travelling, with acceleration 0.3m(s^-2) up a straight road inclined 12 degrees to the horizontal. There is a force resisting the motion of 250 N. Calculate the magnitude of the driving force.
Please help me out on this one. I'm really confused.
Plus, if you have great resources that can help a layman understand Resolving Forces to its depth, please add them too.
Thanks in Advance
classical-mechanics
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
$endgroup$
I have exams after a few days and I'm doing all I can to understand the concept of resolving forces. With hard luck and a few hours of devotion, I acquired basic knowledge on Resolving Forces and was able to solve almost all questions and then this one came out.
A car of mass 850 kg is travelling, with acceleration 0.3m(s^-2) up a straight road inclined 12 degrees to the horizontal. There is a force resisting the motion of 250 N. Calculate the magnitude of the driving force.
Please help me out on this one. I'm really confused.
Plus, if you have great resources that can help a layman understand Resolving Forces to its depth, please add them too.
Thanks in Advance
classical-mechanics
classical-mechanics
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
asked May 6 '16 at 16:46
Kanchan SharmaKanchan Sharma
84
84
$begingroup$
Is the force resisting the motion in addition to gravity?
$endgroup$
– DJohnM
May 6 '16 at 18:20
add a comment |
$begingroup$
Is the force resisting the motion in addition to gravity?
$endgroup$
– DJohnM
May 6 '16 at 18:20
$begingroup$
Is the force resisting the motion in addition to gravity?
$endgroup$
– DJohnM
May 6 '16 at 18:20
$begingroup$
Is the force resisting the motion in addition to gravity?
$endgroup$
– DJohnM
May 6 '16 at 18:20
add a comment |
2 Answers
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If that was actually the way the question was asked? I don't see that the angle is at all relevant- you are told that there is "a force resisting the motion 250 N". Since "F= ma", if the 850 kg car is has 0.3 m/s^2 acceleration, the net force on the car is (850)(0.3)= 255 N. The "driving force" must be that plus the 250 N resisting force it must overcome to move at all,
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
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add a comment |
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As @user247327 has calculated, the net force, acting up the slope must be $255.0 text{ Newtons}$
The question remaining is to determine all the real forces acting along the slope. One of these is the "driving force", $F_D$= and it must be big enough to give the desired net result.
There is a force resisting the motion: $250 text{ Newtons}$. Since the motion is up the slope, this force must be down the slope.
The vertical force of gravity can be resolved into two components: one parallel to the slope and one perpendicular (normal) to the slope. A diagram will show that the parallel component of gravity, $F_P$, is given by:$$F_p= mgsin(12^0)=850 times 9.8times0.207912=1732 text{ Newtons} text{ down the slope}$$
So, taking forces up the slope as positive, we're left with solving:$$255=F_D+(-250)+(-1732)$$
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
$begingroup$
If that was actually the way the question was asked? I don't see that the angle is at all relevant- you are told that there is "a force resisting the motion 250 N". Since "F= ma", if the 850 kg car is has 0.3 m/s^2 acceleration, the net force on the car is (850)(0.3)= 255 N. The "driving force" must be that plus the 250 N resisting force it must overcome to move at all,
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
If that was actually the way the question was asked? I don't see that the angle is at all relevant- you are told that there is "a force resisting the motion 250 N". Since "F= ma", if the 850 kg car is has 0.3 m/s^2 acceleration, the net force on the car is (850)(0.3)= 255 N. The "driving force" must be that plus the 250 N resisting force it must overcome to move at all,
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
$endgroup$
add a comment |
$begingroup$
If that was actually the way the question was asked? I don't see that the angle is at all relevant- you are told that there is "a force resisting the motion 250 N". Since "F= ma", if the 850 kg car is has 0.3 m/s^2 acceleration, the net force on the car is (850)(0.3)= 255 N. The "driving force" must be that plus the 250 N resisting force it must overcome to move at all,
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
$endgroup$
If that was actually the way the question was asked? I don't see that the angle is at all relevant- you are told that there is "a force resisting the motion 250 N". Since "F= ma", if the 850 kg car is has 0.3 m/s^2 acceleration, the net force on the car is (850)(0.3)= 255 N. The "driving force" must be that plus the 250 N resisting force it must overcome to move at all,
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
answered May 6 '16 at 17:42
user247327user247327
11.6k1516
11.6k1516
add a comment |
add a comment |
$begingroup$
As @user247327 has calculated, the net force, acting up the slope must be $255.0 text{ Newtons}$
The question remaining is to determine all the real forces acting along the slope. One of these is the "driving force", $F_D$= and it must be big enough to give the desired net result.
There is a force resisting the motion: $250 text{ Newtons}$. Since the motion is up the slope, this force must be down the slope.
The vertical force of gravity can be resolved into two components: one parallel to the slope and one perpendicular (normal) to the slope. A diagram will show that the parallel component of gravity, $F_P$, is given by:$$F_p= mgsin(12^0)=850 times 9.8times0.207912=1732 text{ Newtons} text{ down the slope}$$
So, taking forces up the slope as positive, we're left with solving:$$255=F_D+(-250)+(-1732)$$
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
$endgroup$
add a comment |
$begingroup$
As @user247327 has calculated, the net force, acting up the slope must be $255.0 text{ Newtons}$
The question remaining is to determine all the real forces acting along the slope. One of these is the "driving force", $F_D$= and it must be big enough to give the desired net result.
There is a force resisting the motion: $250 text{ Newtons}$. Since the motion is up the slope, this force must be down the slope.
The vertical force of gravity can be resolved into two components: one parallel to the slope and one perpendicular (normal) to the slope. A diagram will show that the parallel component of gravity, $F_P$, is given by:$$F_p= mgsin(12^0)=850 times 9.8times0.207912=1732 text{ Newtons} text{ down the slope}$$
So, taking forces up the slope as positive, we're left with solving:$$255=F_D+(-250)+(-1732)$$
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
$endgroup$
add a comment |
$begingroup$
As @user247327 has calculated, the net force, acting up the slope must be $255.0 text{ Newtons}$
The question remaining is to determine all the real forces acting along the slope. One of these is the "driving force", $F_D$= and it must be big enough to give the desired net result.
There is a force resisting the motion: $250 text{ Newtons}$. Since the motion is up the slope, this force must be down the slope.
The vertical force of gravity can be resolved into two components: one parallel to the slope and one perpendicular (normal) to the slope. A diagram will show that the parallel component of gravity, $F_P$, is given by:$$F_p= mgsin(12^0)=850 times 9.8times0.207912=1732 text{ Newtons} text{ down the slope}$$
So, taking forces up the slope as positive, we're left with solving:$$255=F_D+(-250)+(-1732)$$
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
$endgroup$
As @user247327 has calculated, the net force, acting up the slope must be $255.0 text{ Newtons}$
The question remaining is to determine all the real forces acting along the slope. One of these is the "driving force", $F_D$= and it must be big enough to give the desired net result.
There is a force resisting the motion: $250 text{ Newtons}$. Since the motion is up the slope, this force must be down the slope.
The vertical force of gravity can be resolved into two components: one parallel to the slope and one perpendicular (normal) to the slope. A diagram will show that the parallel component of gravity, $F_P$, is given by:$$F_p= mgsin(12^0)=850 times 9.8times0.207912=1732 text{ Newtons} text{ down the slope}$$
So, taking forces up the slope as positive, we're left with solving:$$255=F_D+(-250)+(-1732)$$
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
answered May 6 '16 at 18:51
DJohnMDJohnM
3,2241815
3,2241815
add a comment |
add a comment |
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$begingroup$
Is the force resisting the motion in addition to gravity?
$endgroup$
– DJohnM
May 6 '16 at 18:20