Given:
W = mg = 50N
the component parallel to the plane surface is
mg sin 37 = 50 sin 37 = 30.09N
A block weighing 50 N rests on an inclined plane. It’s weight force directed vertically downward....
36. Ablock weighing 50 N rests on an inclined plane. Its weight is a force directed vertically downward, as illus- trated in v Fig. 3.30. Find the components of the force par allel to the surface of the plane and perpendicular to it. w (50 N) FIGURE 3.30 Block on an inclined plane See Exercise 36. 37
A loaded penguin sled weighing 64 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.20, and the coefficient of kinetic friction is 0.17. (a) What is the minimum magnitude of the force , parallel to the plane, that will prevent the sled from slipping down the plane? ___N (b) What is the minimum magnitude F that will start the sled moving up the plane? ___...
A loaded penguin sled weighing 80 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.26, and the coefficient of kinetic friction is 0.17.What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?What is the minimum magnitude F that will start the sled moving up the plane?What value of F is required to...
A 79 N block rests on a 34° inclined plane. Determine the normal force and find the friction force that keeps the block from sliding.
1. A block weighing 50 kN is supported on an inclined plane forming an angle of 25º with the horizontal. A force P that is parallel to the inclined plane is applied to the block, as shown below. Determine the value of P required that will cause impending motion up the inclined plane, if: a) (10 pts) the coefficient of friction between the body and the plane is 0.30. b) (10 pts) the coefficient of friction between the body and...