A 50-kg crate is pulled across a smooth floor using a rope. The crate is accelerating at 1 m/s $2$ and the rope makes an angle of 50° with the floor. The normal force is
A) 430 N.
B) 540 N.
C) 390 N.
D) 490 N.

Question

Quizplus · Accepted Answer

The normal force can be calculated by balancing the forces in the vertical direction. The weight of the crate acts downwards (W = mg = 50kg * 9.8m/s^2 = 490N) and the vertical component of the tension in the rope acts upwards and reduces the effective weight. The vertical component of the tension (T_y) is T*sin(50°), where T is the tension in the rope. However, we don't need to calculate T to find the normal force. The normal force (N) is equal to the weight minus the vertical component of the tension. Since there's no vertical acceleration, N + T_y = W. But without the exact value of T, we directly address the normal force as being equal to the weight of the crate when there's no vertical acceleration and no other vertical forces are mentioned. Thus, the normal force is equal to the gravitational force acting on the crate, which is 490N, minus the vertical component of the pulling force, which is not directly given but is implied to reduce the normal force. To correct my initial explanation, without the tension's vertical component's exact impact provided or calculated, we cannot directly conclude the normal force without considering the effect of the pulling force's vertical component. Given the information, we cannot accurately calculate the normal force without knowing the tension in the rope. My initial statement was incorrect as it oversimplified the scenario by not accounting for the vertical component of the pulling force correctly.

A 50-Kg Crate Is Pulled Across a Smooth Floor Using 222

A 50-Kg Crate Is Pulled Across a Smooth Floor Using $2$