How Do You Calculate the Tension in a Sleigh Rope with Friction and an Angle?

[Peterr13]

An adult is pulling two small children in a sleigh over level snow. The sleigh and children have a total mass of 47 kg. The sleigh rope makes an angle of 23 degrees with the horizontal. The coeffcient of friction between the sleigh and the snow is 0.11. Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity. (Hint: The normal force is not equal to the force of gravity.)

i don't know how to find force normal

please help me

 

[TheSwager]

The normal is always perpendicular to friction and friction is equal to the coefficient of friction times the normal. another thing which might help is that the y component of the tension vector plus the normal is equal to mg(force on sled by earth)

 

[tiny-tim]

i don't know how to find force normal

Hi Peterr13!

Easy … just take components in the normal direction (even if the velocity is not constant, or the ground is not horizontal, this still works, because the normal acceleration is always zero )

 

[Peterr13]

thanks guys for your help... i'l try it out and see what happens

 

[Peterr13]

so will it be like mg = fn + force applied in y dir
460.6 = 460.60 + sin23(y dir)

 

[tiny-tim]

No, to find f_n take components in the normal direction …

what are the components of the weight the tension and the friction force in the normal direction?

 

[Peterr13]

well isn;t the normal force which 460.6 +the applied force in y dir?

 

[Rasalhague]

so will it be like mg = fn + force applied in y dir
460.6 = 460.60 + sin23(y dir)

Nearly... There's no acceleration in the y direction (vertical to the ground); the sleigh is only moving in the x direction (horizontally). So all the forces in the y direction add up to zero.

$$F_{N} + T \, sin(23 \, \mathrm{degrees}) - mg = 0$$

where $$F_{N}$$ is the magnitude of the normal force, and $$T$$ is the magnitude of the tension vector. The upward force due to the tension is the vertical component (y component) of the tension vector is $$T \, sin(23 \, \mathrm{degrees})$$. These forces both act upward in the opposite direction to the gravitational force, mg which acts downward, hence the difference in sign. So

$$F_{N} + T \, sin(23 \, \mathrm{degrees}) = mg$$

$$\therefore \: F_{N} = mg - T \, sin(23 \, \mathrm{degrees})$$

mg = 460.6 Newtons, but $$T \, sin(23 \, \mathrm{degrees})$$ is not equal to zero--there is some vertical component to the tension--so the magnitude of the normal force will be less than 460.6 Newtons.

 

[Peterr13]

Nearly... There's no acceleration in the y direction (vertical to the ground); the sleigh is only moving in the x direction (horizontally). So all the forces in the y direction add up to zero.

$$F_{N} + T \, sin(23 \, \mathrm{degrees}) - mg = 0$$

where $$F_{N}$$ is the magnitude of the normal force, and $$T$$ is the magnitude of the tension vector. The upward force due to the tension is the vertical component (y component) of the tension vector is $$T \, sin(23 \, \mathrm{degrees})$$. These forces both act upward in the opposite direction to the gravitational force, mg which acts downward, hence the difference in sign. So

$$F_{N} + T \, sin(23 \, \mathrm{degrees}) = mg$$

$$\therefore \: F_{N} = mg - T \, sin(23 \, \mathrm{degrees})$$

mg = 460.6 Newtons, but $$T \, sin(23 \, \mathrm{degrees})$$ is not equal to zero--there is some vertical component to the tension--so the magnitude of the normal force will be less than 460.6 Newtons.

thanks for the reply but the problem i have now is that there is two unknowns so do i substitute one formula to another?