The block A in Fig. P-544 supports a load W and is to be raised by forcing the wedge B under it. If the angle of friction is 10° at all surfaces in contact, determine the maximum wedge angle α that will give the wedge a mechanical advantage; i.e., make P less than the weight W of the block.
To adjust the vertical position of a column supporting 200-kN load, two 5° wedges are used as shown in Fig. P-543. Determine the force P necessary to start the wedges is the angle of friction at all contact surfaces is 25°. Neglect friction at the rollers.
As shown in Fig. P-540, two blocks each weighing 20 kN and resting on a horizontal surface, are to be pushed apart by a 30° wedge. The angle of friction is 15° for all contact surfaces. What value of P is required to start movement of the blocks? How would this answer be changed if the weight of one of the blocks were increased by 30 kN?
The block A in Fig. P-539 supports a load W = 100 kN and is to be raised by forcing the wedge B under it. The angle of friction for all surfaces in contact is f = 15°. If the wedge had a weight of 40 kN, what value of P would be required (a) to start the wedge under the block and (b) to pull the wedge out from under the block?
in Fig. P-536, determine the minimum weight of block B that will keep it at rest while a force P starts blocks A up the incline surface of B. The weight of A is 100 lb and the angle of friction for all surfaces in contact is 15°.
A uniform bar AB, weighing 424 N, is fastened by a frictionless pin to a block weighing 200 N as shown in Fig. P-533. At the vertical wall, μ = 0.268 while under the block, μ = 0.20. Determine the force P needed to start motion to the right.