The net force on the object will be the difference between the two forces, so the net force will be 5 units. Given that the boy is applying 10 units of force in one direction and the girl is applying 5 units of force in the opposite direction, the net force will be 10 units - 5 units = 5 units.
The net force on the crate would be 10 units (force applied by the boy) - 10 units (force applied by the girl) = 0 units. Since the forces are in opposite directions, they cancel each other out, resulting in a net force of zero.
The net force on the crate is 70 N to the left while it's moving to the left, and 20 N to the right while it's moving to the right. If the 100N force and the 50N force are acting on it at the same time, then the net force on the crate is 20 N towards the left.
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.
You should have friction between the box and the ground, the force you used to push it, the mass of the object, and gravity. The resistance of the air is neglibible. This should be all the forces on an object on flat ground.
The net force on the object would be 5 units in the direction of the stronger force, which is the force applied by the boy. This is because the net force is calculated by subtracting the smaller force (5 units) from the larger force (10 units).
The net force on the crate would be 10 units (force applied by the boy) - 10 units (force applied by the girl) = 0 units. Since the forces are in opposite directions, they cancel each other out, resulting in a net force of zero.
crate- direct object
The net force on the crate is 70 N to the left while it's moving to the left, and 20 N to the right while it's moving to the right. If the 100N force and the 50N force are acting on it at the same time, then the net force on the crate is 20 N towards the left.
The direction of friction on the crate is opposite to the direction in which it is sliding. In this case, since you are pushing the crate to the right, the friction will act to the left in order to oppose the motion.
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.
You should have friction between the box and the ground, the force you used to push it, the mass of the object, and gravity. The resistance of the air is neglibible. This should be all the forces on an object on flat ground.
The net force on the object would be 5 units in the direction of the stronger force, which is the force applied by the boy. This is because the net force is calculated by subtracting the smaller force (5 units) from the larger force (10 units).
direct object
If the crate is moving at a constant velocity, then the force of friction acting on the crate is equal in magnitude and opposite in direction to the force you are applying to push the crate. This means that the force you apply to push the crate is balancing out the force of friction acting against it. By measuring the force you are exerting and observing the constant velocity of the crate, you can infer the magnitude of the friction force.
The force that resists the motion of the crate is the force of friction between the crate and the floor. This frictional force acts in the opposite direction to the pushing force applied by the person, making it harder to move the crate.
If the crate isn't accelerating ... i.e. sliding at a constant speed, not speeding up or slowing down ...then the forces on it are balanced. The pseudo-force of friction is 140N in the direction opposite toits speed.
Direct Object