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 the drag due to air resistance is negligible, what are the total north-south net forces and the total east-west net force acting on the crate, causing it to accelerate across the floor?(1 point)

Question

Responses

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

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

795 N to the south and 850 N to the west
795 N to the south and 850 N to the west

855 N to the South and 759 N and to the West

Answer 1

To find the total net forces acting on the crate, we need to calculate the net forces in the north-south direction and the east-west direction separately.

North-South Forces:

• The force from the worker pushing north: \( +960 , \text{N} \)
• The force of friction acting south: \( -105 , \text{N} \)

Net north-south force: \[ F_{NS} = 960 , \text{N} - 105 , \text{N} = 855 , \text{N} , \text{(to the north)} \]

East-West Forces:

• The force from the worker pushing east: \( +875 , \text{N} \)
• The force of friction acting west: \( -80 , \text{N} \)

Net east-west force: \[ F_{EW} = 875 , \text{N} - 80 , \text{N} = 795 , \text{N} , \text{(to the east)} \]

Summary of Forces:

• Total north-south net force: \( 855 , \text{N} , \text{(to the north)} \)
• Total east-west net force: \( 795 , \text{N} , \text{(to the east)} \)

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