To give solution to the exercise we must use the concepts of Torque, Vector magnitude and vector direction of the forces.
For the given problem we have to
In this way the torque acting on the particle as a function of distance and time is,
The net torque acting on the particle is
PART B) The direction of the torque is given by,
Therefore the torque direction is 48.04° below the x axis.
Answer:8.968 N-m
Explanation:
Given
Length of arm=0.152 m
Downward force=118 N
angle made by arm with vertical
Force can be divided into two components
It's sin component will contribute towards torque while cos component will not contibute
T=8.968 N-m
a.
The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:
where F is the magnitude of the force, theta is the angle between them and d is the distance.
The problen gives the following data:
The magnitude of the force 750 N.
The angle between the force and the displacement which is 25°
The distance, 26 m.
Plugging this in the formula we have:
Therefore the work done is 17673 J.
b)
The power is given by:
the problem states that the time it takes is 6 s. Then:
Therefore the power is 2945.5 W