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8. Block B weighs 710 N. The coefficient of static friction between the block and horizontal surface is 0.25. Calculate the maximum weight of block A for which the system will be in equilibrium. I mut: 127.5

8. Block B weighs 710 N. The coefficient of static friction between the block and horizontal surface is 0.25. Calculate the maximum weight of block A for which the system will be in equilibrium. I mut: 127.5

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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7. A block of mass m sits on the ground. A student pulls up on the block with a tension T, but the block remains in contact with the ground. What is the normal force on the block? 1775- (mg) = 0 $8 Fg = mg. 8. Block B weighs 710 N. The coefficient of static friction between the block and horizontal surface is 0.25. Calculate the maximum. weight of block A for which the system will be in equilibrium. Smax(25) (710)=177.5N T₁ must = 177.5 mg21775 N m T₁=177.5N 9. Picture given below shows the motion of two boxes under the effect of applied force. Friction constant between the surfaces is k-0.4. Find the acceleration of the boxes and tension on the rope. T+F-10m=0 = 710N m₂=1kg F₁ = mg.-T N £<>T VF₁ = 710N B 10. When system is in motion, find the acceleration and the tension on the rope. می تر ₂-177.50 T₁ mi-3kg F=20N T=? A 37° 12258 T₁-172 177 45° F=30NT
Transcribed Image Text:7. A block of mass m sits on the ground. A student pulls up on the block with a tension T, but the block remains in contact with the ground. What is the normal force on the block? 1775- (mg) = 0 $8 Fg = mg. 8. Block B weighs 710 N. The coefficient of static friction between the block and horizontal surface is 0.25. Calculate the maximum. weight of block A for which the system will be in equilibrium. Smax(25) (710)=177.5N T₁ must = 177.5 mg21775 N m T₁=177.5N 9. Picture given below shows the motion of two boxes under the effect of applied force. Friction constant between the surfaces is k-0.4. Find the acceleration of the boxes and tension on the rope. T+F-10m=0 = 710N m₂=1kg F₁ = mg.-T N £<>T VF₁ = 710N B 10. When system is in motion, find the acceleration and the tension on the rope. می تر ₂-177.50 T₁ mi-3kg F=20N T=? A 37° 12258 T₁-172 177 45° F=30NT
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