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Two workers are pushing on the same large crate. One worker pushes east with a force of 875 N, while the other worker pushes north with a force of 960 N. Friction between the crate and the floor is directed west with a force of 80 N and south with a force of 105 N. Assuming drag due to air resistance is negligible, what are the total north-south net forces and the total east-west net forces acting on the crate, causing it to accelerate across the floor?(1 point)
Responses

795 N to the north and 855 N to the east
795 N to the north and 855 N to the east

855 N to the south and 759 N to the west
855 N to the south and 759 N to the west

795 N to the south and 855 N to the west
795 N to the south and 855 N to the west

855 N to the north and 795 N to the east
855 N to the north and 795 N to the east
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1 answer

To calculate the total net forces acting on the crate, we need to consider the forces in both the north-south and east-west directions.

North-South Forces:

  • Northward Force: 960 N (from worker)
  • Southward Force: 105 N (friction)

The net north-south force can be calculated as: \[ F_{\text{north-south}} = F_{\text{north}} - F_{\text{south}} = 960 , \text{N} - 105 , \text{N} = 855 , \text{N} \text{ (to the north)} \]

East-West Forces:

  • Eastward Force: 875 N (from worker)
  • Westward Force: 80 N (friction)

The net east-west force can be calculated as: \[ F_{\text{east-west}} = F_{\text{east}} - F_{\text{west}} = 875 , \text{N} - 80 , \text{N} = 795 , \text{N} \text{ (to the east)} \]

Results:

  • Total north-south net force: 855 N to the north
  • Total east-west net force: 795 N to the east

Thus, the answer is: 855 N to the north and 795 N to the east.