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If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.
The resultant of the two forces F1 and F2 is
with direction at
north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be 
south-east.
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.
The resultant of the two forces F1 and F2 is
with direction at
Answer:
The kinetic coefficient of friction of the crate is 0.235.
Explanation:
As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:
(Ec. 1)
(Ec. 2)
Where:
- Pushing force, measured in newtons.
- Tension, measured in newtons.
- Coefficient of kinetic friction, dimensionless.
- Normal force, measured in newtons.
- Weight of the crate, measured in newtons.
The system of equations is now reduced by algebraic means:
And we finally clear the coefficient of kinetic friction and apply the definition of weight:
If we know that ,
,
and
, then:
The kinetic coefficient of friction of the crate is 0.235.
