Question
A nurse pushes a cart by exerting a force on the handle at a downward angle 35.035.0^\circ below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity?
Question by OpenStax is licensed under CC BY 4.0
Final Answer

73.2 N73.2 \textrm{ N}

Solution video

OpenStax College Physics, Chapter 4, Problem 35 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We have a free body diagram of the cart being pushed by the nurse. It consists of an applied force at an angle of 35 degrees below the horizontal and has friction force going straight back 60 newtons. There’s a normal force going straight up and gravity going down. The mass of the cart is 28 kilograms and its acceleration is zero because we’re told that it’s going to be pushed at constant speed. So the net force in the x direction is the x component of this applied force minus the friction and that’s going to be mass times acceleration but we know that acceleration is zero so we replace that with a zero on the right side. On the left side, we’ll replace Fax with the applied force times cos theta because Fax is the adjacent leg of this triangle here. Ultimately we’re trying to find out what is this applied force needed to move at constant speed. So we’ll add friction to both sides and then divide both sides by cos theta, and we get the applied force is friction divided by cos theta, so that’s 60 newtons divided by cos 35 which is 73.2 newtons.

Comments

Why is the downward force only gravity and not the weight of the cart?

Hi jsotosky, are you also asking "Why is the downward force only gravity and not the weight of the cart?"
The force of gravity on the cart, which is downward, is the weight. The only other force with a downward component is the force applied by the nurse, which is illustrated as a force applied at 35.035.0^\circ below horizontal.
Does that answer the question?
All the best,
Shaun