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)

Question

Responses

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

795 N to the north and 855 N to the east

Answer 1

To determine the total net forces acting on the crate, we need to analyze the forces in both the north-south (y-axis) and east-west (x-axis) directions.

Forces in the North-South Direction:

• Worker 1 pushes north with a force of 960 N.
• The frictional force acts south with a force of 105 N.

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

Forces in the East-West Direction:

• Worker 2 pushes east with a force of 875 N.
• The frictional force acts west with a force of 80 N.

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

Putting the net forces together, we find:

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

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