I have no clue what the normal force is :-(

[HelloMotto]

An adult is pulling two small children in sleigh over level snow. The sleigh and the children have a total mass of 47 kg. The sleigh rope makes an angle of 23 with the horizontal, The coefficient of kinetic 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 constant speed.

The book gives me a hint saying that the normal force is not equal in magnitude of the gravity.

I don't get why they are not equal...because the sleigh is still on a leveled surface... so why is not same as Fg? And how do i figure out the normal force?

 

[Wecht]

That was a pretty stupid hint for the book to give, so I'm going to assume that you read the question incorrectly or mistyped. Say, the book says the normal force is not equal in magnitude to the weight. That makes sense (so did the other hint, but it was unnecessary).

You have to account for the upwards force of the slanted pull.
Fp(vector) = Fp(cos)~ x-direction + Fp(sin)~ y-direction
where Fp=force of pull and ~=angle with respect to the horizontal

So, Fw (weight vector) is equal in magnitude to the combined magnitudes of Fn (normal force) and Fp(sin)~
Fw - [Fn + Fp(sin)~] = 0 N

 

[baba89]

The normal force is the perpendicular force exerted by a surface on an object in contact with it. In this case, the normal force is the force exerted by the snow on the sleigh. It is not equal to the force of gravity because the surface of the snow is not flat and level. The snow will deform under the weight of the sleigh and children, creating a normal force that is less than the force of gravity.

To calculate the magnitude of the tension in the rope needed to keep the sleigh moving at constant speed, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration. In this case, the net force is the tension in the rope minus the force of friction. The acceleration is zero since the sleigh is moving at a constant speed.

We can set up an equation: T - μN = 0, where T is the tension in the rope, μ is the coefficient of kinetic friction, and N is the normal force. We can solve for T by substituting the given values:

T - (0.11)(47 kg)(9.8 m/s^2) = 0
T = 51.2 N

Therefore, the magnitude of the tension in the rope needed to keep the sleigh moving at constant speed is 51.2 N.