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What is the weight of a 15kg crate resting on a 30.0 degree incline?
Wiki User
∙ 12y ago
Weight is a force, so you use the equation F=Ma
Fw=15(9.81)
Fw=147.15
Fw=150 N
Wiki User
∙ 12y ago
The weight of the crate can be broken into two components: one parallel to the incline and one perpendicular to the incline. Using trigonometry, you can find that the weight component parallel to the incline is Wsin(30°) and the weight component perpendicular to the incline is Wcos(30°), where W is the weight of the crate.
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If a crate weighing 508 N is resting on a plane inclined 26degrees above the horizontal if acceleration is 4.29 then how fast will the crate be moving after 6 seconds?
To determine the speed of the crate after 6 seconds, we first need to calculate the net force acting on the crate on the inclined plane. This can be done by resolving the weight of the crate into components parallel and perpendicular to the plane. Then, using Newton's second law, F = ma, where F is the net force, m is the mass of the crate, and a is the acceleration, we can find the acceleration down the incline. After finding this acceleration, we can use the kinematic equation v = u + at to calculate the final speed of the crate after 6 seconds, where v is the final velocity, u is the initial velocity (assumed to be 0), a is the acceleration, and t is the time.
When a crate slide down an incline at a constant velocity it is?
When a crate slides down an incline at a constant velocity, it is experiencing a balanced force situation. The force of gravity pulling it downhill is counteracted by the force of friction acting in the opposite direction. This results in the crate moving steadily without speeding up or slowing down.
What is the force of friction acting on a crate that slides across the floor?
The force of friction acting on a crate sliding across the floor is equal in magnitude but opposite in direction to the force applied to move the crate. It depends on the coefficient of friction between the crate and the floor, as well as the weight of the crate.
A plane inclined at an angle 35 degrees from horizontal if the coefficient of static friction between the crate and the plane is 0.65 will the crate slide down the plane?
Is mgsinΘ > μmgcosΘ ? Is sinΘ > μ cosΘ ? Is sin35º > .65 cos35º Is .573 > .532 => Yes, so crate slides down the plane, no matter what the mass is or acceleration due to gravity
A 33 kg crate rests on a horizontal floor and a 58 kg person is standing on the crate Determine the magnitude of the normal force that the floor exerts on the crate.?
The normal force the floor exerts on the crate is equal in magnitude and opposite in direction to the weight of both the crate and the person standing on it. Therefore, the normal force is equal to the sum of the weight of the crate (33 kg * 9.8 m/s^2) and the weight of the person (58 kg * 9.8 m/s^2). Calculate the total weight and that will give you the magnitude of the normal force exerted by the floor, which is 33 kg * 9.8 m/s^2 + 58 kg * 9.8 m/s^2.
If someone wants to slid a 100 pound crate up an incline plan to a loading dock 4 feet above the groundand the incline is 20 feet long how much effort will it take to slid th crate up the incline?
Does anyone know the answer for this? If you do, please help!!
If a crate weighing 508 N is resting on a plane inclined 26degrees above the horizontal if acceleration is 4.29 then how fast will the crate be moving after 6 seconds?
To determine the speed of the crate after 6 seconds, we first need to calculate the net force acting on the crate on the inclined plane. This can be done by resolving the weight of the crate into components parallel and perpendicular to the plane. Then, using Newton's second law, F = ma, where F is the net force, m is the mass of the crate, and a is the acceleration, we can find the acceleration down the incline. After finding this acceleration, we can use the kinematic equation v = u + at to calculate the final speed of the crate after 6 seconds, where v is the final velocity, u is the initial velocity (assumed to be 0), a is the acceleration, and t is the time.
A crate is at rest on the ground what force or forces are acting on the crate?
The weight of the crate is acting downward on the ground and the ground is exerting a force equal to the weight of the crate upward on the crate.
When a crate slide down an incline at a constant velocity it is?
When a crate slides down an incline at a constant velocity, it is experiencing a balanced force situation. The force of gravity pulling it downhill is counteracted by the force of friction acting in the opposite direction. This results in the crate moving steadily without speeding up or slowing down.
What force or forces are acting on the crate?
The weight of the crate is acting downward on the ground and the ground is exerting a force equal to the weight of the crate upward on the crate.
Is a crate resting on a loading ramp an example of static friction or fluid friction?
me
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