A man pushes on a piano of mass 180 kg so that it slides at a constant velocity of 12.0 cm/s down a ramp that is inclined at 11.0degrees above the horizontal. No appreciable friction is acting on the piano. Calculate the magnitude and direction of this push
(a)if the man pushes parallel to the incline,
We take the upward direction to be positive and downward one to be negative. The sign of the force should be entered correctly.
(b)if the man pushes the piano up the plane instead, also at 12.0 cm/s parallel to the incline,
(c)if the man pushes horizontally, but still with a speed of 12.0 cm/s.
Bobpursley,
That doesnt answer the question for c???
(a) If the man pushes parallel to the incline, the force he exerts will only counteract the component of the weight of the piano that is parallel to the incline.
To calculate the magnitude of the push, we need to find the component of the weight of the piano parallel to the incline. Using trigonometry, this component can be found using the equation:
F = m * g * sin(theta)
Where:
F = force parallel to the incline
m = mass of the piano
g = acceleration due to gravity
theta = angle of inclination
Plugging in the given values:
F = 180 kg * 9.8 m/s^2 * sin(11°)
Calculating this:
F ≈ 320.5 N
So, the magnitude of the push required is approximately 320.5 N.
Since the piano is sliding down the ramp at a constant velocity, the push should have the same magnitude but in the opposite direction. Therefore, the direction of the push is downward and negative.
(b) If the man pushes the piano up the plane instead, the force he exerts should not only counteract the component of the weight parallel to the incline but also provide additional force to oppose the piano's motion.
Since the piano is still sliding down the ramp at a constant velocity, this means the force of gravity is greater than the force the man exerts.
To calculate the magnitude of the push, we need to find the component of the weight of the piano parallel to the incline and subtract it from the force required to oppose the piano's motion.
Using the equation from part (a) to find the component of weight parallel to the incline:
F_parallel = m * g * sin(theta)
Plugging in the given values:
F_parallel = 180 kg * 9.8 m/s^2 * sin(11°)
Calculating this:
F_parallel ≈ 320.5 N
To counteract the piano's motion, the man needs to exert a force equal to the component of weight parallel to the incline plus an additional force equal to the force of gravity acting on the piano. Since the piano's weight is equal to its mass times gravity:
F_push = F_parallel + m * g
Plugging in the given values:
F_push = 320.5 N + 180 kg * 9.8 m/s^2
Calculating this:
F_push ≈ 491.6 N
So, the magnitude of the push required to push the piano up the incline is approximately 491.6 N.
Since the piano is sliding down the ramp at a constant velocity, the push should have the same magnitude but in the opposite direction. Therefore, the direction of the push is downward and negative.
(c) If the man pushes horizontally, but still with a speed of 12.0 cm/s, the force he exerts needs to counteract the component of the weight of the piano that is perpendicular to the incline.
To calculate the magnitude of the push, we need to find the component of the weight of the piano perpendicular to the incline. Using trigonometry, this component can be found using the equation:
F_perpendicular = m * g * cos(theta)
Where:
F_perpendicular = force perpendicular to the incline
m = mass of the piano
g = acceleration due to gravity
theta = angle of inclination
Plugging in the given values:
F_perpendicular = 180 kg * 9.8 m/s^2 * cos(11°)
Calculating this:
F_perpendicular ≈ 1798 N
So, the magnitude of the push required to push the piano horizontally is approximately 1798 N.
Since the piano is sliding down the ramp at a constant velocity, the push should have the same magnitude but in the opposite direction. Therefore, the direction of the push is horizontal, opposite to the direction of motion.
To find the magnitude and direction of the push in each scenario, we need to analyze the forces acting on the piano.
(a) When the man pushes parallel to the incline:
In this case, the force applied by the man will be in the same direction as the motion of the piano. Since the piano is moving at a constant velocity, the net force acting on it must be zero. The force applied by the man will balance the component of the weight acting down the incline, which is given by:
F_push = m * g * sin(theta)
where m is the mass of the piano (180 kg), g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of inclination (11.0 degrees).
Substituting the values:
F_push = 180 kg * 9.8 m/s^2 * sin(11.0 degrees) = 310.47 N
Therefore, the magnitude of the push is 310.47 N, and since it is in the same direction as the motion of the piano, the direction of the push is downward or negative.
(b) When the man pushes the piano up the incline:
In this case, the force applied by the man will be in the opposite direction to the motion of the piano. Since the piano is still moving at a constant velocity, the net force acting on it must be zero. The force applied by the man will need to balance the component of the weight acting down the incline, plus provide an additional force to move the piano up the incline.
F_push = m * g * sin(theta) + m * a
where a is the acceleration of the piano, which is zero because it is moving at a constant velocity. Thus, the force applied by the man will balance only the component of the weight:
F_push = m * g * sin(theta)
Substituting the values:
F_push = 180 kg * 9.8 m/s^2 * sin(11.0 degrees) = 310.47 N
Therefore, the magnitude of the push is 310.47 N, but since it is in the opposite direction to the motion of the piano, the direction of the push is upward or positive.
(c) When the man pushes horizontally:
In this case, the force applied by the man will be perpendicular to the incline, and therefore, it will not affect the motion of the piano up or down the incline. The only force that matters is the gravitational force acting downward, which is balanced by the normal force exerted by the ramp.
Therefore, there is no need for the man to apply any force horizontally since the motion of the piano is already maintained by the equilibrium between gravity and the normal force.
weight has a downward component;
weight in negative direction mg*CosTheta
a) going down the plane, without friction, the man would have to chaase the piano to push it. Have you ever chased a sled on a icy hill?
b) pushing up> Push=frition=mgSinTheta
c) horizontal? then his push in the up direction is push*cosTheta=mg*sinTheta