A block of mass M is pulled along the floor by the force F, shown. The coefficient of kinetic friction is µx. and Hs respectively. What is the frictional force on the block? Don't guess! Use the free body diagram method. Fo O Hk ( Mg - Fo cos(e) ) O Hk (Mg + Fo) O Hk ( Mg + Fo cos(0) ) O Hk (Mg - Fo sin(e) ) Ο μ F sin (θ) O Hk Fo cos(e) O Hk ( Mg - Fo) Ο μμ ( Mg + , si (θ) )|
A block of mass M is pulled along the floor by the force F, shown. The coefficient of kinetic friction is µx. and Hs respectively. What is the frictional force on the block? Don't guess! Use the free body diagram method. Fo O Hk ( Mg - Fo cos(e) ) O Hk (Mg + Fo) O Hk ( Mg + Fo cos(0) ) O Hk (Mg - Fo sin(e) ) Ο μ F sin (θ) O Hk Fo cos(e) O Hk ( Mg - Fo) Ο μμ ( Mg + , si (θ) )|
Related questions
Question
100%
Help, 2 questions
Expert Solution
Step 1
Hello. Since you have posted multiple questions and not specified which question needs to be solved, we will solve the first question for you. If you want any other specific question to be solved then please resubmit only that question or specify the question number.
Answer:
Consider the following free body diagram of the block.
Here, N is the normal reaction force acting on the block.
Here, we have resolved F0 into its components along the X and Y-axes.
Step by step
Solved in 2 steps with 3 images