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PhysicsCanuck

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- Homework Statement
- (a) A man applies a force of F = 300 N at an angle of 60.0° to the door, 2.00 m from well-oiled hinges. Find the torque on the door, choosing the position of the hinges as the axis of rotation.

(b) Suppose a wedge is placed 1.50 m from the hinges on the other side of the door. What minimum force must the wedge exert so that the force applied in part (a) won’t open the door?

My question: Why is the force of the wedge not negative?

My reasoning: The torque is directed clockwise (and would be a negative value), and since r would be positive, then Fsin(theta) should be negative, rendering the torque negative.

- Relevant Equations
- t = rF

t = rF sin(theta)

As per (b) in the above image, or easily solved with t=rFsin(theta), the perpendicular force is 260N.

When inputting that value into the equation for torque, the value for torque is 520Nm, as per t=2.00*300*sin60.

Because the wedge prevents the door from moving, the torque on/at the doorknob (in the counter-clockwise direction) is equal and opposite the torque from/at the wedge, thus the wedge torque is -520Nm.

t_wedge = -520Nm = 1.50*F*sin(theta)

Therefore, F*sin(theta) = -347N

and F = -347N/sin(theta)

Since the provided answer is F = 347, then sin(theta) is equal to -1, and thus theta is equal to -90.

But in my estimation, the wedge force should be

**(negative)**-260N since the opposing force (the perpendicular force pushing counter-clockwise on the door) is

**(positive)**+260N.

Any new way of looking at this and reconciling these opposing forces would be appreciated.

Thank you very much