Q.1 Find the force needed to accelerate a 2 kg block at 3m/s^2 up a rough plane (coefficient of friction 0.2) inclined at 25 degree to the horizontal if the force is
a) Parallel to the slope
b) horizontal
c) at 45 degree to the upward verticle
First force is F which is pushing block upward
Second is the opposing forces, opposite to F
The block accelerates, so there is a net force (UNbalanced forces) which is net F =(F-opposing forces).
Since they are unbalanced and acceleration A results:
netF = m A
So
F-muR-mgsin25=ma
F-0.2mgcos25-mgsin25=ma
F-3.55-8.28=6
F=17.8N
b)
Fcos25-muR-19.6sin25=6.......eq.1
here R=Fsin25+mgcos25 because the force F is horizontal
Fcos25-muR-19.6sin25=6
substitute the value of R in eq.1
Fcos25-mu(Fsin25+mgcos25)-19.6sin25=6
0.906F-0.0845F-3.553-8.28=6
0.821F=17.833
F=17.833/8.21=21.7 N
about part 3 have no clue, i can not figure out the actual angle of force F
3. The angle which the force F makes with the direction of block motion (x-axis) is β=90 - 45 - 25 = 20 degr.
x: ma = F•cos β - m•g•sinα,
y: 0 = -m•g•cosα + N +Fsin β.
F = (m•a + m•g•sinα)/cos β = (2•3+2•9.8•sin25)/cos20 = 15.2 N
thanks a lot Elena, your help is much appreciated
No frictional force in no.3?
To find the force needed to accelerate the block up the rough plane, we need to consider the forces acting on the block and use Newton's second law of motion.
Let's break down the problem into three cases:
a) Force parallel to the slope:
In this case, the force is acting in the same direction as the block's motion. To find the force needed, we'll use the equation:
Force = mass × acceleration
In this case, the acceleration is the same as the given acceleration of 3 m/s^2. So, we have:
Force = 2 kg × 3 m/s^2
Force = 6 N
Therefore, the force needed to accelerate the 2 kg block at 3 m/s^2 up the rough plane when the force is parallel to the slope is 6 N.
b) Horizontal force:
In this case, the force is acting horizontally, perpendicular to the slope. We'll need to find the component of force parallel to the slope and the component perpendicular to the slope.
The force parallel to the slope is given by:
Force_parallel = Force × sin(θ)
where θ is the angle of the slope, which is 25 degrees in this case.
The force perpendicular to the slope is given by:
Force_perpendicular = Force × cos(θ)
The force required to overcome friction is given by:
Force_friction = coefficient_of_friction × force_normal
where the force_normal is the force acting perpendicular to the slope, which is equal to the weight of the block.
The weight of the block is given by:
Weight = mass × gravity
where gravity is 9.8 m/s^2.
So the force_normal is:
Force_normal = Weight × cos(θ)
Force_normal = (2 kg × 9.8 m/s^2) × cos(25 degrees)
The force required to overcome friction is:
Force_friction = 0.2 × Force_normal
Now we can find the required force:
Force = Force_parallel + Force_friction
c) Force at 45 degrees to the upward vertical:
In this case, we'll need to find the component of force parallel to the slope and the component perpendicular to the slope.
The force parallel to the slope is given by:
Force_parallel = Force × sin(θ)
The force perpendicular to the slope is given by:
Force_perpendicular = Force × cos(θ)
We'll also need to find the force required to overcome friction using the same formula as in case b:
Force_friction = coefficient_of_friction × force_normal
Again, the force required to accelerate the block is given by:
Force = Force_parallel + Force_friction
By following these steps, we can calculate the force needed for each case.