Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity.
A man pulls two children in a sleigh over level snow. The sleigh and the children have a total weight of 47 N. The sleigh rope makes an angle of 23 degrees with the horizontal. The coefficient of kinetic friction between the sleigh and the snow is 0.11.
The sleigh has to be kept moving at a constant velocity. This is possible if the force applied by the man who is pulling the sleigh is equal to the force due to friction in the opposite direction.
The force of friction is given Fc*N, where Fc is the coefficient of kinetic friction and N is the normal force.
The rope makes an angle of 23 degrees with the horizontal. If the tension in the rope is T, we can divide it into two components. The vertical component acting upwards is equal to T*sin 23. The normal force of the sleigh is 47 - T*sin 23. This gives the frictional force as (47 - T*sin 23)*0.11
The horizontal component of the tension is pulling the sleigh and is equal to T*cos 23. This is equal to the force of friction.
T*cos 23 = (47 - T*sin 23)*0.11
=> T*cos 23 = 47*0.11 - T*sin 23*0.11
=> T(cos 23 + 0.11*sin 23) = 47*0.11
=> T = 47*0.11 / (cos 23 + 0.11*sin 23)
=> T = 5.356 N
The required tension in the rope is 5.356 N