Question
Expert Solution
arrow_forward
Step 1
Given:
The applied force is .
The mass of the block is .
The acceleration due to gravity is .
The dimension of the block is .
Step by stepSolved in 2 steps
Knowledge Booster
Similar questions
- 2. A uniform ladder weighing 330. N is leaning against a wall. The ladder slips when the angle with the floor is 50.00. Assuming the coefficient of static friction at the wall and the floor are the same, obtain a value for us.arrow_forwardIt is sometimes important to determine the location of a person’s center of gravity(which coincides with the center of mass because the gravitational force is uniform at the scale of the human body). This can be done with the arrangement shown intheFigure below. A light (negligible mass) plank rests on two scales that read Fg1=380 N and Fg2=320 N. The scales are separated by a distance of 2.00 m. How far from the woman’s feet is her center of gravity?arrow_forward2. A 3.0-m-long ladder leans against a wall at an angle of 60° with respect to the floor. What is the minimum value of us, the coefficient of static friction with the ground, that will prevent the ladder from slipping? Assume that friction between the ladder and the wall is negligible. Center of gravity 19 TULO L 60⁰ Weight acts at the center of gravity. Known L-30 m Find Me Contro <= 0 about this point. Static friction prevents slipping.arrow_forward
- A 1.5 kg block resting ina rough flat surface is pulled by a string. The string makes an angle of 20 degress with the horizontal. The coefficient of static friction between the blodk and the surface is 0.35. What is the maximum tension in the string to keep the block unmoved?arrow_forwardProblem A block of mass m is placed on a rough surface inclined relative to the horizontal. The incline angle is increased until the block start to move. Show that you can obtain the coefficient of static friction Us by measuring the critical angle e where slipping occurs. Solution Just before slipping, we say that the block is still at rest, so the net force on the x-axis is: EFx=mg Evaluating, we get: = mg But the frictional force is expressed as the product of coefficient and the normal force so. = mg For an inclined plane. The nomal force is n=mg Then, Hs=mg cos(e) so: tan(e)arrow_forwardNiloarrow_forward
arrow_back_ios
arrow_forward_ios