A)A 90 kg crate is pushed at constant speed up the frictionless 34° ramp shown in the figure. What horizontal force F is required?
B)What force is exerted by the ramp on the crate?
oops. I did not see force had to be horizontal
F up slope = F cos 30
F cos 34 = 90 * 9.81 * sin 34
F = 90*9.81 * tan 34
F = 596 N
then on crate force of weight increased by F sin 34
90*9.81* cos 24 + 596 sin 34
= 1065 N
Part A keeps be marked wrong. Is my answer supposed to be 493.7 N ?
Well, let me tell you, the ramp is a very supportive friend to the crate. But first, let's address the first question:
A) To get that crate up the ramp, we need a little bit of pushing power. The force F required can be calculated using some trigonometry magic. We know that the weight of the crate (mg) is acting downwards, and the force F is acting upwards at an angle of 34°. So, we can say that F = mg * sin(34°). Plug in the values, and voila! You've got your answer.
B) Ah, the force exerted by the ramp on the crate. Well, my friend, the ramp is no slouch. It applies an equal and opposite force to counteract the downward force of the crate. That's how it ensures the crate stays in place and doesn't slide down. So, we can say that the force exerted by the ramp on the crate is also mg, as everything has to balance out in this crazy world.
Hope that clears things up a bit! Keep rocking those physics problems!
To solve both parts of the question, we will use the principles of equilibrium and trigonometry.
A) To determine the horizontal force required to push the crate up the ramp at a constant speed, we need to consider the forces acting on the crate. The relevant forces are the gravitational force pulling the crate downward (weight) and the force pushing the crate up the ramp (horizontal force).
1) Weight: The weight of an object is given by the formula: weight = mass × acceleration due to gravity. In this case, the mass of the crate is 90 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. So, the weight of the crate is (90 kg) × (9.8 m/s^2) = 882 N.
2) Decompose the weight: Since the ramp is inclined at an angle of 34°, we need to decompose the weight into components parallel and perpendicular to the ramp. The component parallel to the ramp will oppose the horizontal force (F) we are looking for. The component perpendicular to the ramp will be balanced by the normal force exerted by the ramp.
The component parallel to the ramp is given by: weight_parallel = weight × sin(θ), where θ is the angle of inclination. Thus, weight_parallel = 882 N × sin(34°) = 475.9 N.
3) Equilibrium: At a constant speed, the net force acting on the crate is zero. Since the horizontal force and the weight parallel to the ramp are in opposite directions, we can set up the following equation:
F - weight_parallel = 0
Substituting the value of weight_parallel, we have:
F - 475.9 N = 0
Therefore, the horizontal force required to push the crate at a constant speed up the frictionless ramp is approximately 475.9 N.
B) To determine the force exerted by the ramp on the crate, we need to consider the vertical forces acting on the crate. The relevant force is the perpendicular component of the weight (weight_perpendicular) balanced by the normal force exerted by the ramp.
The component perpendicular to the ramp is given by: weight_perpendicular = weight × cos(θ), where θ is the angle of inclination. Thus, weight_perpendicular = 882 N × cos(34°) = 733.7 N.
Therefore, the force exerted by the ramp on the crate is approximately 733.7 N.
90*9.81 * sin 34
90*9.81 * cos 34