A 74 Newton toolbox is dragged horiztonally at constant speed by a rope making a

Question

A 74 Newton toolbox is dragged horiztonally at constant speed by a rope making an angle of 20 degrees with the floor. The tension in the rope is 43 newtons. Determine the magnitude of the friction force and the normal force.

Friction Force =  [Num]Newtons

Normal Force =  [Num]Newtons

A 74 Newton toolbox is dragged horiztonally at constant speed by a rope making an angle of 20 degrees with the floor. The tension in the rope is 43 newtons. Determine the magnitude of the friction force and the normal force.

Friction Force =  [Num]Newtons

Normal Force =  [Num]Newtons

Explanation / Answer

A 74 Newton toolbox is dragged horiztonally at constant speed by a rope making an angle of 20 degrees with the floor. The tension in the rope is 43 newtons. Determine the magnitude of the friction force and the normal force.

So, you need to look at all of the forces in this problem. Horizontal forces do not effect vertical forces and vertical forces do not effect horizontal forces, so treat them separately. And when you have a force at an angle as you do, break it up into horizontal and vertical component forces using simple trig.

Vertical forces:
Well, you have the weight force of the box, which is 74 N. Because the box is not flying into the air or sinking into the ground, this 74N force down needs to be countered with a 74N force up.
In this case, you have two upward forces. The upward force that is a result of the rope being pulled at an angle, and the normal force which is the force of the ground pushing up on the box so that it does not sink into the ground. To find the upward force as a result of the rope, think of it as a triangle. In this case, the upward component force would be 43sin20. The normal force is the variable you will be solving for.

Sum of all downward forces = Sum of all upward forces
74 N = (Fn) + 43sin20
Solve for Fn to find normal force
Fn= 59.29 N

Horizontal Forces:
You have 2 Horizontal forces here. The horizontal force created from the x-component of the rope pull, and the force of friction.
ONLY because the box is going at a constant speed (as in not being accelerated) can we say that the 2 forces must be equal.
So, again, look at the rope at a 20 deg angle as a triangle, and you get that the horizontal component is 43cos20. So, the frictional force must also be 43cos20

Hope this helped, if these are not the right answers or you would like further explanation please comment to let me know

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